Calculate indefinite integral $\int{ \frac{x}{(x+1)^3}dx }$

Question

I want to calculate the following integral by using the table of immediate integrals (no integral substitution):

$$\int{ \frac{x}{(x+1)^3}dx }$$

I took this formula from the immediate integral table of my textbook:

$$\int{f'(x)\ [f(x)]^\alpha\ dx} = \frac {1}{\alpha + 1}[f(x)]^{\alpha+1}+C$$

So this is what I do:

$$\int{ x \ \frac{1}{(x+1)^3}dx } = \int{ x \ (x+1)^{-3} dx }$$

In my integral $\alpha = -3$, $f(x) = x + 1$ and $f'(x) = x + 1$. Therefore

$$\int{ x \ (x+1)^{-3} dx } = { -\frac{1}{2(x+1)^2} dx\ + C}$$

The solution, though, is the following:

$${-\frac{1}{x+1} + \frac{1}{2(x+1)^2} dx\ + C}$$

This clearly doesn't match with my solution, although it's quite similar. Any hints on what I am doing wrong?

Answer 1

You don't have to use those tables.

(You should be using $u$-substitution; I'm essentially doing the same.)

$$\begin{array}{rcl} \displaystyle \int \dfrac x{(x+1)^3} \ \mathrm dx &=& \displaystyle \int \dfrac {(x+1)-1}{(x+1)^3} \ \mathrm dx \\ &=& \displaystyle \int \left( \dfrac1{(x+1)^2} - \dfrac1{(x+1)^3} \right) \ \mathrm dx \\ &=& \displaystyle \int \left( \dfrac1{(x+1)^2} - \dfrac1{(x+1)^3} \right) \ \mathrm d(x+1) \\ &=& \displaystyle - \dfrac1{x+1} + \dfrac1{2(x+1)^2} + C\\ \end{array}$$

Answer 2

Here's another fun way:

$$\begin{align}\int\frac x{(1+x)^{n+1}}\ \mathrm dx&=-\frac1n\frac\partial{\partial t}\int\frac1{(1+xt)^n}\ \mathrm dx\bigg|_{t=1}\\&=\frac1{n(n-1)}\frac\partial{\partial t}\frac1{t(1+xt)^{n-1}}\bigg|_{t=1}\\&=-\frac1{n(n-1)}\frac{xt+nt+1-t}{t^2(1+xt)^n}\bigg|_{t=1}\\&=-\frac{x+n}{n(n-1)(1+x)^n}\end{align}$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int{x \over \pars{x + 1}^{3}}\,\dd x & = \int{1/x^{2} \over \pars{1 + 1/x}^{3}}\,\dd x = -\int{1 \over \pars{1 + 1/x}^{3}}\,\dd\pars{1 \over x} = {1 \over 2\pars{1 + 1/x}^{2}} \\[5mm] & = \bbx{\ds{{x^{2} \over 2\pars{x + 1}^{2}} + \pars{~\mbox{a constant}~}}} \end{align}

Different answers can show "different" solutions. However, with a "little algebra" it's shown that the difference of two of them is a constant $\ds{\pars{~x\mbox{-}independent~}}$ !!!.