Integral check. Is partial fractions the only way?Integral by partial fractionsHow to use integration by...
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Integral check. Is partial fractions the only way?
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Integral check. Is partial fractions the only way?
Integral by partial fractionsHow to use integration by partial fractions to calculate the integral of a function whose numerator is larger than its denominator?Find the integral of $frac{sin(x)}{1+sin^2(x)}$Integral Partial Fractions and Wolfram AlphaResolve into partial fractions, hence calculate the volume obtainedPartial fractions to solve integralsSolve $int frac{1}{cos^2(x)+cos(x)+1}dx$Evaluate $int frac{e^{2x}-1}{e^{2x}+1}dx$ using partial fractionscomplex integral, partial fractions 2Integral using trig substitution check
$begingroup$
I'm getting a different answer from wolfram and I have no idea where. I have to integrate:
$$int_0^1 frac{xdx}{(2x+1)^3}$$
Is partial fractions the only way?
So evaluating the fraction first:
$$frac{x}{(2x+1)^3} = frac{A}{2x+1} + frac{B}{(2x+1)^2} + frac{C}{(2x+1)^3}$$
$$x = A(2x+1)^2 + B(2x+1) + C$$
$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$
$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$
$$x = x^2(4A) + x(4A+2B) + A + B + C$$
$4A = 0$ and $4A+2B = 1$ and $A + B + C = 0$ so $A = 0$ and $B=frac{1}{2}$ and $C = frac{-1}{2}$
Is the partial fraction part right?
So then I get:
$$int_0^1 frac{xdx}{(2x+1)^3} = int_0^1 frac{dx}{(2(2x+1)^2)} - int_0^1 frac{dx}{2(2x+1)^3}$$
for both, using $u = 2x+1$ and $frac{du}{dx} = 2$ and $du = 2dx$ and $frac{du}{2} = dx$,
$$frac{1}{4} int frac{du}{u^2} - frac{1}{4} int frac{du}{u^3}$$
$$ = [frac{1}{4} - u^{-1} - frac{1}{4} cdot frac{1}{-2} u^{-2} ]_1^3$$
I'm going the route of changing the limits to the new u and since $u = 2x+1$, when $x = 0, u = 1$ and when $x = 1, u = 3$. Is this the right path?
finally I get
$$[frac{-1}{4}u^{-1} + frac{1}{8}u^{-2} ]_1^3$$
I plug in numbers but I get a different answer than wolfram...
integration
edited 2 hours ago
Joshua Mundinger
2,5191028
asked 2 hours ago
Jwan622Jwan622
2,13611530
$endgroup$
add a comment |
$begingroup$
I'm getting a different answer from wolfram and I have no idea where. I have to integrate:
$$int_0^1 frac{xdx}{(2x+1)^3}$$
Is partial fractions the only way?
So evaluating the fraction first:
$$frac{x}{(2x+1)^3} = frac{A}{2x+1} + frac{B}{(2x+1)^2} + frac{C}{(2x+1)^3}$$
$$x = A(2x+1)^2 + B(2x+1) + C$$
$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$
$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$
$$x = x^2(4A) + x(4A+2B) + A + B + C$$
$4A = 0$ and $4A+2B = 1$ and $A + B + C = 0$ so $A = 0$ and $B=frac{1}{2}$ and $C = frac{-1}{2}$
Is the partial fraction part right?
So then I get:
$$int_0^1 frac{xdx}{(2x+1)^3} = int_0^1 frac{dx}{(2(2x+1)^2)} - int_0^1 frac{dx}{2(2x+1)^3}$$
for both, using $u = 2x+1$ and $frac{du}{dx} = 2$ and $du = 2dx$ and $frac{du}{2} = dx$,
$$frac{1}{4} int frac{du}{u^2} - frac{1}{4} int frac{du}{u^3}$$
$$ = [frac{1}{4} - u^{-1} - frac{1}{4} cdot frac{1}{-2} u^{-2} ]_1^3$$
I'm going the route of changing the limits to the new u and since $u = 2x+1$, when $x = 0, u = 1$ and when $x = 1, u = 3$. Is this the right path?
finally I get
$$[frac{-1}{4}u^{-1} + frac{1}{8}u^{-2} ]_1^3$$
I plug in numbers but I get a different answer than wolfram...
integration
edited 2 hours ago
Joshua Mundinger
2,5191028
asked 2 hours ago
Jwan622Jwan622
2,13611530
$endgroup$
$begingroup$
I can't see any mistake in your work. What's the answer given by wolfram?
$endgroup$
– Thomas Shelby
2 hours ago
$begingroup$
They have 1/18 as the answer @ThomasShelby
$endgroup$
– Jwan622
39 mins ago
add a comment |
$begingroup$
I'm getting a different answer from wolfram and I have no idea where. I have to integrate:
$$int_0^1 frac{xdx}{(2x+1)^3}$$
Is partial fractions the only way?
So evaluating the fraction first:
$$frac{x}{(2x+1)^3} = frac{A}{2x+1} + frac{B}{(2x+1)^2} + frac{C}{(2x+1)^3}$$
$$x = A(2x+1)^2 + B(2x+1) + C$$
$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$
$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$
$$x = x^2(4A) + x(4A+2B) + A + B + C$$
$4A = 0$ and $4A+2B = 1$ and $A + B + C = 0$ so $A = 0$ and $B=frac{1}{2}$ and $C = frac{-1}{2}$
Is the partial fraction part right?
So then I get:
$$int_0^1 frac{xdx}{(2x+1)^3} = int_0^1 frac{dx}{(2(2x+1)^2)} - int_0^1 frac{dx}{2(2x+1)^3}$$
for both, using $u = 2x+1$ and $frac{du}{dx} = 2$ and $du = 2dx$ and $frac{du}{2} = dx$,
$$frac{1}{4} int frac{du}{u^2} - frac{1}{4} int frac{du}{u^3}$$
$$ = [frac{1}{4} - u^{-1} - frac{1}{4} cdot frac{1}{-2} u^{-2} ]_1^3$$
I'm going the route of changing the limits to the new u and since $u = 2x+1$, when $x = 0, u = 1$ and when $x = 1, u = 3$. Is this the right path?
finally I get
$$[frac{-1}{4}u^{-1} + frac{1}{8}u^{-2} ]_1^3$$
I plug in numbers but I get a different answer than wolfram...
integration
edited 2 hours ago
Joshua Mundinger
2,5191028
asked 2 hours ago
Jwan622Jwan622
2,13611530
$endgroup$
I'm getting a different answer from wolfram and I have no idea where. I have to integrate:
$$int_0^1 frac{xdx}{(2x+1)^3}$$
Is partial fractions the only way?
So evaluating the fraction first:
$$frac{x}{(2x+1)^3} = frac{A}{2x+1} + frac{B}{(2x+1)^2} + frac{C}{(2x+1)^3}$$
$$x = A(2x+1)^2 + B(2x+1) + C$$
$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$
$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$
$$x = x^2(4A) + x(4A+2B) + A + B + C$$
$4A = 0$ and $4A+2B = 1$ and $A + B + C = 0$ so $A = 0$ and $B=frac{1}{2}$ and $C = frac{-1}{2}$
Is the partial fraction part right?
So then I get:
$$int_0^1 frac{xdx}{(2x+1)^3} = int_0^1 frac{dx}{(2(2x+1)^2)} - int_0^1 frac{dx}{2(2x+1)^3}$$
for both, using $u = 2x+1$ and $frac{du}{dx} = 2$ and $du = 2dx$ and $frac{du}{2} = dx$,
$$frac{1}{4} int frac{du}{u^2} - frac{1}{4} int frac{du}{u^3}$$
$$ = [frac{1}{4} - u^{-1} - frac{1}{4} cdot frac{1}{-2} u^{-2} ]_1^3$$
I'm going the route of changing the limits to the new u and since $u = 2x+1$, when $x = 0, u = 1$ and when $x = 1, u = 3$. Is this the right path?
finally I get
$$[frac{-1}{4}u^{-1} + frac{1}{8}u^{-2} ]_1^3$$
I plug in numbers but I get a different answer than wolfram...
integration
integration
edited 2 hours ago
Joshua Mundinger
2,5191028
asked 2 hours ago
Jwan622Jwan622
2,13611530
edited 2 hours ago
Joshua Mundinger
2,5191028
asked 2 hours ago
Jwan622Jwan622
2,13611530
edited 2 hours ago
Joshua Mundinger
2,5191028
edited 2 hours ago
Joshua Mundinger
2,5191028
edited 2 hours ago
Joshua Mundinger
2,5191028
2,5191028
asked 2 hours ago
Jwan622Jwan622
2,13611530
asked 2 hours ago
Jwan622Jwan622
2,13611530
asked 2 hours ago
Jwan622Jwan622
2,13611530
2,13611530
$begingroup$
I can't see any mistake in your work. What's the answer given by wolfram?
$endgroup$
– Thomas Shelby
2 hours ago
$begingroup$
They have 1/18 as the answer @ThomasShelby
$endgroup$
– Jwan622
39 mins ago
add a comment |
$begingroup$
I can't see any mistake in your work. What's the answer given by wolfram?
$endgroup$
– Thomas Shelby
2 hours ago
$begingroup$
They have 1/18 as the answer @ThomasShelby
$endgroup$
– Jwan622
39 mins ago
$begingroup$
I can't see any mistake in your work. What's the answer given by wolfram?
$endgroup$
– Thomas Shelby
2 hours ago
$begingroup$
I can't see any mistake in your work. What's the answer given by wolfram?
$endgroup$
– Thomas Shelby
2 hours ago
$begingroup$
They have 1/18 as the answer @ThomasShelby
$endgroup$
– Jwan622
39 mins ago
$begingroup$
They have 1/18 as the answer @ThomasShelby
$endgroup$
– Jwan622
39 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: Substitute $x=tfrac12 u -tfrac12$
answered 2 hours ago
MPWMPW
30.3k12157
$endgroup$
$begingroup$
You beat me to it.
$endgroup$
– randomgirl
2 hours ago
$begingroup$
If I go your way, I wind up with $frac{1}{2} int frac{1}{u^2} - frac{1}{2} int frac{1}{u^3}$ which is different than what I have at some point. I can't see my error.
$endgroup$
– Jwan622
16 mins ago
add a comment |
$begingroup$
A much much easier way to solve it is by using integration by parts.
Hint: Take $u=x$, $dv = frac{dx}{(2x+1)^3}$.
answered 2 hours ago
Haris GusicHaris Gusic
960116
$endgroup$
$begingroup$
$u$ but then $dv$?
$endgroup$
– manooooh
2 hours ago
$begingroup$
@manooooh What do you mean? I don't understand.
$endgroup$
– Haris Gusic
2 hours ago
$begingroup$
I am sorry, I understood that you used sub. My apologies.
$endgroup$
– manooooh
2 hours ago
add a comment |
$begingroup$
1) Observe: $$x= frac 12 times left ( (2x + 1) -1right).$$
2) Use this to obtain $$frac{x}{(2x+1)^3} = frac{1}{2} times frac{1}{(2x+1)^2} - frac 12 times frac{1}{(2x+1)^3}.$$
3) Integrate to get
$$ int frac{x}{(2x+1)^3} dx = -frac{1}{4} (2x+1)^{-1} + frac{1}{8} (2x+1)^{-2}.$$
answered 2 hours ago
FnacoolFnacool
5,021511
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Substitute $x=tfrac12 u -tfrac12$
answered 2 hours ago
MPWMPW
30.3k12157
$endgroup$
$begingroup$
You beat me to it.
$endgroup$
– randomgirl
2 hours ago
$begingroup$
If I go your way, I wind up with $frac{1}{2} int frac{1}{u^2} - frac{1}{2} int frac{1}{u^3}$ which is different than what I have at some point. I can't see my error.
$endgroup$
– Jwan622
16 mins ago
add a comment |
$begingroup$
Hint: Substitute $x=tfrac12 u -tfrac12$
answered 2 hours ago
MPWMPW
30.3k12157
$endgroup$
$begingroup$
You beat me to it.
$endgroup$
– randomgirl
2 hours ago
$begingroup$
If I go your way, I wind up with $frac{1}{2} int frac{1}{u^2} - frac{1}{2} int frac{1}{u^3}$ which is different than what I have at some point. I can't see my error.
$endgroup$
– Jwan622
16 mins ago
add a comment |
$begingroup$
Hint: Substitute $x=tfrac12 u -tfrac12$
answered 2 hours ago
MPWMPW
30.3k12157
$endgroup$
Hint: Substitute $x=tfrac12 u -tfrac12$
answered 2 hours ago
MPWMPW
30.3k12157
answered 2 hours ago
MPWMPW
30.3k12157
answered 2 hours ago
MPWMPW
30.3k12157
answered 2 hours ago
MPWMPW
30.3k12157
30.3k12157
$begingroup$
You beat me to it.
$endgroup$
– randomgirl
2 hours ago
$begingroup$
If I go your way, I wind up with $frac{1}{2} int frac{1}{u^2} - frac{1}{2} int frac{1}{u^3}$ which is different than what I have at some point. I can't see my error.
$endgroup$
– Jwan622
16 mins ago
add a comment |
$begingroup$
You beat me to it.
$endgroup$
– randomgirl
2 hours ago
$begingroup$
If I go your way, I wind up with $frac{1}{2} int frac{1}{u^2} - frac{1}{2} int frac{1}{u^3}$ which is different than what I have at some point. I can't see my error.
$endgroup$
– Jwan622
16 mins ago
$begingroup$
You beat me to it.
$endgroup$
– randomgirl
2 hours ago
$begingroup$
You beat me to it.
$endgroup$
– randomgirl
2 hours ago
$begingroup$
If I go your way, I wind up with $frac{1}{2} int frac{1}{u^2} - frac{1}{2} int frac{1}{u^3}$ which is different than what I have at some point. I can't see my error.
$endgroup$
– Jwan622
16 mins ago
$begingroup$
If I go your way, I wind up with $frac{1}{2} int frac{1}{u^2} - frac{1}{2} int frac{1}{u^3}$ which is different than what I have at some point. I can't see my error.
$endgroup$
– Jwan622
16 mins ago
add a comment |
$begingroup$
A much much easier way to solve it is by using integration by parts.
Hint: Take $u=x$, $dv = frac{dx}{(2x+1)^3}$.
answered 2 hours ago
Haris GusicHaris Gusic
960116
$endgroup$
$begingroup$
$u$ but then $dv$?
$endgroup$
– manooooh
2 hours ago
$begingroup$
@manooooh What do you mean? I don't understand.
$endgroup$
– Haris Gusic
2 hours ago
$begingroup$
I am sorry, I understood that you used sub. My apologies.
$endgroup$
– manooooh
2 hours ago
add a comment |
$begingroup$
A much much easier way to solve it is by using integration by parts.
Hint: Take $u=x$, $dv = frac{dx}{(2x+1)^3}$.
answered 2 hours ago
Haris GusicHaris Gusic
960116
$endgroup$
$begingroup$
$u$ but then $dv$?
$endgroup$
– manooooh
2 hours ago
$begingroup$
@manooooh What do you mean? I don't understand.
$endgroup$
– Haris Gusic
2 hours ago
$begingroup$
I am sorry, I understood that you used sub. My apologies.
$endgroup$
– manooooh
2 hours ago
add a comment |
$begingroup$
A much much easier way to solve it is by using integration by parts.
Hint: Take $u=x$, $dv = frac{dx}{(2x+1)^3}$.
answered 2 hours ago
Haris GusicHaris Gusic
960116
$endgroup$
A much much easier way to solve it is by using integration by parts.
Hint: Take $u=x$, $dv = frac{dx}{(2x+1)^3}$.
answered 2 hours ago
Haris GusicHaris Gusic
960116
answered 2 hours ago
Haris GusicHaris Gusic
960116
answered 2 hours ago
Haris GusicHaris Gusic
960116
answered 2 hours ago
Haris GusicHaris Gusic
960116
960116
$begingroup$
$u$ but then $dv$?
$endgroup$
– manooooh
2 hours ago
$begingroup$
@manooooh What do you mean? I don't understand.
$endgroup$
– Haris Gusic
2 hours ago
$begingroup$
I am sorry, I understood that you used sub. My apologies.
$endgroup$
– manooooh
2 hours ago
add a comment |
$begingroup$
$u$ but then $dv$?
$endgroup$
– manooooh
2 hours ago
$begingroup$
@manooooh What do you mean? I don't understand.
$endgroup$
– Haris Gusic
2 hours ago
$begingroup$
I am sorry, I understood that you used sub. My apologies.
$endgroup$
– manooooh
2 hours ago
$begingroup$
$u$ but then $dv$?
$endgroup$
– manooooh
2 hours ago
$begingroup$
$u$ but then $dv$?
$endgroup$
– manooooh
2 hours ago
$begingroup$
@manooooh What do you mean? I don't understand.
$endgroup$
– Haris Gusic
2 hours ago
$begingroup$
@manooooh What do you mean? I don't understand.
$endgroup$
– Haris Gusic
2 hours ago
$begingroup$
I am sorry, I understood that you used sub. My apologies.
$endgroup$
– manooooh
2 hours ago
$begingroup$
I am sorry, I understood that you used sub. My apologies.
$endgroup$
– manooooh
2 hours ago
add a comment |
$begingroup$
1) Observe: $$x= frac 12 times left ( (2x + 1) -1right).$$
2) Use this to obtain $$frac{x}{(2x+1)^3} = frac{1}{2} times frac{1}{(2x+1)^2} - frac 12 times frac{1}{(2x+1)^3}.$$
3) Integrate to get
$$ int frac{x}{(2x+1)^3} dx = -frac{1}{4} (2x+1)^{-1} + frac{1}{8} (2x+1)^{-2}.$$
answered 2 hours ago
FnacoolFnacool
5,021511
$endgroup$
add a comment |
$begingroup$
1) Observe: $$x= frac 12 times left ( (2x + 1) -1right).$$
2) Use this to obtain $$frac{x}{(2x+1)^3} = frac{1}{2} times frac{1}{(2x+1)^2} - frac 12 times frac{1}{(2x+1)^3}.$$
3) Integrate to get
$$ int frac{x}{(2x+1)^3} dx = -frac{1}{4} (2x+1)^{-1} + frac{1}{8} (2x+1)^{-2}.$$
answered 2 hours ago
FnacoolFnacool
5,021511
$endgroup$
add a comment |
$begingroup$
1) Observe: $$x= frac 12 times left ( (2x + 1) -1right).$$
2) Use this to obtain $$frac{x}{(2x+1)^3} = frac{1}{2} times frac{1}{(2x+1)^2} - frac 12 times frac{1}{(2x+1)^3}.$$
3) Integrate to get
$$ int frac{x}{(2x+1)^3} dx = -frac{1}{4} (2x+1)^{-1} + frac{1}{8} (2x+1)^{-2}.$$
answered 2 hours ago
FnacoolFnacool
5,021511
$endgroup$
1) Observe: $$x= frac 12 times left ( (2x + 1) -1right).$$
2) Use this to obtain $$frac{x}{(2x+1)^3} = frac{1}{2} times frac{1}{(2x+1)^2} - frac 12 times frac{1}{(2x+1)^3}.$$
3) Integrate to get
$$ int frac{x}{(2x+1)^3} dx = -frac{1}{4} (2x+1)^{-1} + frac{1}{8} (2x+1)^{-2}.$$
answered 2 hours ago
FnacoolFnacool
5,021511
answered 2 hours ago
FnacoolFnacool
5,021511
answered 2 hours ago
FnacoolFnacool
5,021511
answered 2 hours ago
FnacoolFnacool
5,021511
5,021511
add a comment |
add a comment |
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$begingroup$
I can't see any mistake in your work. What's the answer given by wolfram?
$endgroup$
– Thomas Shelby
2 hours ago
$begingroup$
They have 1/18 as the answer @ThomasShelby
$endgroup$
– Jwan622
39 mins ago