Htydjtk

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...

$$int_0^1 frac{xdx}{(2x+1)^3} =
int_0^1 frac{dx}{2(2x+1)^2} - int_0^1 frac{dx}{2(2x+1)^3}$$

$displaystyle int dfrac{dx}{2(2x+1)^2}:$

$qquad u=2x+1, x=dfrac{u-1}{2}, dx = dfrac 12 du$

$qquad displaystyle int dfrac{dx}{2(2x+1)^2} =
int dfrac{du}{4u^2} = -dfrac{1}{4u}$

$qquad x=0 mapsto u=1, x=1 mapsto u=3$

$qquad left[-dfrac{1}{4u} right]_1^3 = -dfrac{1}{12} + dfrac 14 = dfrac 16$

$displaystyle int dfrac{dx}{2(2x+1)^3}:$

$qquad u=2x+1, x=dfrac{u-1}{2}, dx = dfrac 12 du$

$qquad displaystyle int dfrac{dx}{2(2x+1)^3} =
int dfrac{du}{4u^3} = -dfrac{1}{8u^2}$

$qquad x=0 mapsto u=1, x=1 mapsto u=3$

$qquad left[-dfrac{1}{8u^2} right]_1^3 = -dfrac{1}{72} + dfrac 18 = dfrac 19$

$$int_0^1 frac{dx}{2(2x+1)^2} - int_0^1 frac{dx}{2(2x+1)^3}
= dfrac 16 - dfrac 19 = dfrac{1}{18}$$

Hint: Substitute $x=tfrac12 u -tfrac12$

Integrand: We have $x=tfrac12 u -tfrac12=tfrac12(u-1)$, so
$$frac{x}{(2x+1)^3}=frac{tfrac12(u-1)}{(2cdotfrac12(u-1)+1)^3}
=tfrac12frac{u-1}{u^3}=boxed{tfrac12left(u^{-2}-u^{-3}right)}$$

Differential: We have $x=tfrac12 u -tfrac12$, so
$$dx = dleft(tfrac12 u -tfrac12right)=boxed{tfrac12du}$$

Limits of integration: The limits for $x$ are $0$ and $1$, so we find the corresponding values of $u$:
$$x=0implies 0=tfrac12 u -tfrac12implies u=boxed{1}$$
$$x=1implies 1 = tfrac12 u -tfrac12 implies u=boxed{3}$$
So, replacing the integrand, differential, and limits with the boxed items above we have
$$int_0^1frac{x; dx}{(2x+1)^3} = int_1^3tfrac12left(u^{-2}-u^{-3}right) tfrac12du$$
$$=tfrac14int_1^3left(u^{-2}-u^{-3}right) du$$
$$=left.frac14left(-u^{-1}+tfrac12u^{-2} right)right]_1^3$$
$$=tfrac14[(-tfrac13+tfrac1{18})-(-1+tfrac12)]$$
$$=tfrac14[-tfrac{5}{18}+tfrac12]$$
$$=tfrac14[tfrac{4}{18}]$$
$$=boxed{tfrac{1}{18}}$$

This is the precise procedure you should follow for any substitution. Don't take shortcuts.

You've computed the integral correctly, probably you made a mistake while substituting in the limits of integration. Everything else upto that point is correct.

Is partial fractions the only way?

No single way is the only way, you have tons of ways to do this integral. Have a look at the following approaches.

1. 
   

   Manipulation and U-Substitution

   
   
   
   
   
   • Make the substitution $begin{bmatrix}t \ mathrm dt end{bmatrix}=begin{bmatrix}2x+1 \ 2mathrm dxend{bmatrix}$
     
   
   
   • Use the General Power Rule for Integrals
     $$I=int_{0}^{1}dfrac{x mathrm dx}{(2x+1)^3}implies 2I=int_{0}^{1}dfrac{2x+1-1}{(2x+1)^3}mathrm dx$$ $$2I=int_{0}^{1}biggl[dfrac{1}{(2x+1)^2}-dfrac{1}{(2x+1)^3}biggr] mathrm dx $$
     
   
   
   
   

$$implies I=dfrac{1}{4}biggl[-dfrac{1}{t}+dfrac{1}{2t^2}biggr]_{1}^{3}=dfrac{1}{18}$$

2. 
   

   Trigonometric Substitution

   
   
   
   
   
   • Make the substitution $begin{bmatrix}x \ mathrm dxend{bmatrix}=begin{bmatrix}1/2cdottan^2 theta\tanthetasec^2thetamathrm dthetaend{bmatrix}$
     
   
   
   • Use the Pythagorean Identity involving tangent and secant functions namely $sec^2theta=1+tan^2theta$ to get $2x+1=sec^2theta$.
   
   
   
   

$$I=int_{0}^{1}dfrac{xmathrm dx}{(2x+1)^3}=int_{0}^{sqrt{2}}dfrac{tan^3theta sec^2theta}{sec^6theta}mathrm dtheta$$

Using the definitions of $tantheta=sintheta/costheta$ and $sectheta=1/costheta$. The integral simplifies to the following form: $$I=int_{0}^{sqrt{2}}sin^3thetacosthetamathrm dtheta=int_{0}^{sqrt{2}}sin^3theta cdotmathrm d(sintheta)$$

Again use the General Power Formula for Integrals to get the following expression:

$$I=dfrac{1}{8}biggl[sin^4tan^{-1}(sqrt{2x})biggr]_{0}^{1}=dfrac{1}{18}$$

To compute $sinarctan(sqrt{2})$, make use of the following easy to prove identity: $$sintan^{-1} x=dfrac{x}{sqrt{1+x^2}}$$

A much much easier way to solve it is by using integration by parts.

Hint: Take $u=x$, $dv = frac{dx}{(2x+1)^3}$.

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

$$ int frac{x}{(2x+1)^3} dx = -frac{1}{4} (2x+1)^{-1} + frac{1}{8} (2x+1)^{-2}.$$