Integral check. Is partial fractions the only way? 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...
 A: Hint: Substitute $x=\tfrac12 u -\tfrac12$

Addendum: Your original problem was $\int_0^1\frac{x\; dx}{(2x+1)^3}$. For this substitution, compute:

Integrand: We have $x=\tfrac12 u -\tfrac12=\tfrac12(u-1)$, so
$$\frac{x}{(2x+1)^3}=\frac{\tfrac12(u-1)}{(2\cdot\frac12(u-1)+1)^3}
=\tfrac12\frac{u-1}{u^3}=\boxed{\tfrac12\left(u^{-2}-u^{-3}\right)}$$
Differential: We have $x=\tfrac12 u -\tfrac12$, so
$$dx = d\left(\tfrac12 u -\tfrac12\right)=\boxed{\tfrac12du}$$
Limits of integration: The limits for $x$ are $0$ and $1$, so we find the corresponding values of $u$:
$$x=0\implies 0=\tfrac12 u -\tfrac12\implies u=\boxed{1}$$
$$x=1\implies 1 = \tfrac12 u -\tfrac12 \implies u=\boxed{3}$$
So, replacing the integrand, differential, and limits with the boxed items above we have
$$\int_0^1\frac{x\; dx}{(2x+1)^3} = \int_1^3\tfrac12\left(u^{-2}-u^{-3}\right) \tfrac12du$$
$$=\tfrac14\int_1^3\left(u^{-2}-u^{-3}\right) du$$
$$=\left.\frac14\left(-u^{-1}+\tfrac12u^{-2} \right)\right]_1^3$$
$$=\tfrac14[(-\tfrac13+\tfrac1{18})-(-1+\tfrac12)]$$
$$=\tfrac14[-\tfrac{5}{18}+\tfrac12]$$
$$=\tfrac14[\tfrac{4}{18}]$$
$$=\boxed{\tfrac{1}{18}}$$
This is the precise procedure you should follow for any substitution. Don't take shortcuts. 
A: 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.


*

*Manipulation and U-Substitution


*

*Make the substitution $\begin{bmatrix}t \\ \mathrm dt \end{bmatrix}=\begin{bmatrix}2x+1 \\ 2\mathrm dx\end{bmatrix}$

*Use the General Power Rule for Integrals
$$I=\int_{0}^{1}\dfrac{x \mathrm dx}{(2x+1)^3}\implies 2I=\int_{0}^{1}\dfrac{2x+1-1}{(2x+1)^3}\mathrm dx$$ $$2I=\int_{0}^{1}\biggl[\dfrac{1}{(2x+1)^2}-\dfrac{1}{(2x+1)^3}\biggr] \mathrm dx $$
$$\implies I=\dfrac{1}{4}\biggl[-\dfrac{1}{t}+\dfrac{1}{2t^2}\biggr]_{1}^{3}=\dfrac{1}{18}$$



*Trigonometric Substitution


*

*Make the substitution $\begin{bmatrix}x \\ \mathrm dx\end{bmatrix}=\begin{bmatrix}1/2\cdot\tan^2 \theta\\\tan\theta\sec^2\theta\mathrm d\theta\end{bmatrix}$

*Use the Pythagorean Identity involving tangent and secant functions namely $\sec^2\theta=1+\tan^2\theta$ to get $2x+1=\sec^2\theta$.



$$I=\int_{0}^{1}\dfrac{x\mathrm dx}{(2x+1)^3}=\int_{0}^{\sqrt{2}}\dfrac{\tan^3\theta \sec^2\theta}{\sec^6\theta}\mathrm d\theta$$
Using the definitions of $\tan\theta=\sin\theta/\cos\theta$ and $\sec\theta=1/\cos\theta$. The integral simplifies to the following form: $$I=\int_{0}^{\sqrt{2}}\sin^3\theta\cos\theta\mathrm d\theta=\int_{0}^{\sqrt{2}}\sin^3\theta \cdot\mathrm d(\sin\theta)$$
Again use the General Power Formula for Integrals to get the following expression:
$$I=\dfrac{1}{8}\biggl[\sin^4\tan^{-1}(\sqrt{2x})\biggr]_{0}^{1}=\dfrac{1}{18}$$
To compute $\sin\arctan(\sqrt{2})$, make use of the following easy to prove identity: $$\sin\tan^{-1} x=\dfrac{x}{\sqrt{1+x^2}}$$
A: A much much easier way to solve it is by using  integration by parts.
Hint: Take $u=x$, $dv = \frac{dx}{(2x+1)^3}$.
A: 1) Observe:  $$x= \frac 12 \times \left ( (2x + 1) -1\right).$$ 
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 
$$ \int \frac{x}{(2x+1)^3} dx  = -\frac{1}{4} (2x+1)^{-1} + \frac{1}{8} (2x+1)^{-2}.$$
A: $$\int_0^1 \frac{xdx}{(2x+1)^3} = 
  \int_0^1 \frac{dx}{2(2x+1)^2} - \int_0^1 \frac{dx}{2(2x+1)^3}$$
$\displaystyle \int \dfrac{dx}{2(2x+1)^2}:$
$\qquad u=2x+1,\ x=\dfrac{u-1}{2}, \ dx = \dfrac 12 du$
$\qquad \displaystyle \int \dfrac{dx}{2(2x+1)^2} = 
                      \int \dfrac{du}{4u^2} = -\dfrac{1}{4u}$
$\qquad x=0 \mapsto u=1, \ x=1 \mapsto u=3$
$\qquad \left[-\dfrac{1}{4u} \right]_1^3 = -\dfrac{1}{12} + \dfrac 14 = \dfrac 16$
$\displaystyle \int \dfrac{dx}{2(2x+1)^3}:$
$\qquad u=2x+1,\ x=\dfrac{u-1}{2}, \ dx = \dfrac 12 du$
$\qquad \displaystyle \int \dfrac{dx}{2(2x+1)^3} = 
                      \int \dfrac{du}{4u^3} = -\dfrac{1}{8u^2}$
$\qquad x=0 \mapsto u=1, \ x=1 \mapsto u=3$
$\qquad \left[-\dfrac{1}{8u^2} \right]_1^3 = -\dfrac{1}{72} + \dfrac 18 = \dfrac 19$
$$\int_0^1 \frac{dx}{2(2x+1)^2} - \int_0^1 \frac{dx}{2(2x+1)^3}
   = \dfrac 16 - \dfrac 19 = \dfrac{1}{18}$$
There is also another way to solve for $A, B$, and $C$.
$$\frac{A}{2x+1} + \frac{B}{(2x+1)^2} + \frac{C}{(2x+1)^3} =\frac{x}{(2x+1)^3}$$
$$A(2x+1)^2 + B(2x+1) + C = x$$
Let $x = -\dfrac 12$ and you get
\begin{align}
   C &= -\dfrac 12 \\
   A(2x+1)^2 + B(2x+1) - \dfrac 12 &= x \\
   2A(2x+1)^2 + 2B(2x+1) &= 2x+1 \\
   2A(2x+1) + 2B &= 1
\end{align}
Again, let $x = -\dfrac 12$ and you get
\begin{align}
   2B &= 1 \\
   B &= \dfrac 12 \\
   2A(2x+1) + 1 &= 1 \\
   A &= 0
\end{align}
A: There are many ways to evaluate this integral. One of them is to consider the function
$$I(n;a,b)=I=\int_0^1\frac{xdx}{(ax+b)^n}$$
for $b>0$, $0\neq a>-1$, and $1,2\neq n>0$. First we multiply the RHS by $1=a/a$:
$$I=\frac1a\int_0^1\frac{axdx}{(ax+b)^n}=\frac1a\int_0^1\frac{ax+b-b}{(ax+b)^n}dx$$
$$I=\frac1a\int_0^1\frac{dx}{(ax+b)^{n-1}}-\frac{b}a\int_0^1\frac{dx}{(ax+b)^n}$$
Then we focus on $$J(n;a,b)=J=\int_0^1\frac{dx}{(ax+b)^n}$$
Sub.: $$u=ax+b\Rightarrow \frac{du}a=dx$$
$$x=1\mapsto u=b+a\\ x=0\mapsto u=b$$
thus
$$J=\frac1a\int_{b}^{b+a}\frac{du}{u^n}=\frac1{a(1-n)}\left[(b+a)^{1-n}-b^{1-n}\right]$$
And since 
$$I(n;a,b)=\frac1aJ(n-1;a,b)-\frac{b}aJ(n;a,b)$$
We have that
$$I(n;a,b)=\frac1{a^2}\left[\frac{(b+a)^{2-n}-b^{2-n}}{2-n}-b\frac{(b+a)^{1-n}-b^{1-n}}{n-1}\right]$$
And of course your integral is given by $I(3;2,1)=1/18$. 
No partial fractions needed :)
PS: This even works for fractional values of $n$. Check it out here.
