Integral check. Is partial fractions the only way?
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.
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 much much easier way to solve it is by using integration by parts.
Hint: Take $u=x$, $dv = \frac{dx}{(2x+1)^3}$.