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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?

Cesare
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    Someone thought $x$ is the derivative of $x+1$ – DHMO Apr 17 '17 at 09:39
  • Thanks @DHMO! I am now trying to figure out how to actually calculate the integral of that by looking at the immediate integrals table. Would appreciate any hints. Feel free to post your comment as an answer and I will be happy to accept it since you solved my problem. – Cesare Apr 17 '17 at 09:48
  • $\alpha$ in your example is $-\frac{3}{2}$ instead of $3$. Further you need a factor of $\frac{1}{2}$ since the derivative of $f$ is $2x$. – Piotr Benedysiuk Apr 17 '17 at 09:57

3 Answers3

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You don't have to use those tables.

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

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

DHMO
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Here's another fun way:

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

Simply Beautiful Art
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $\ds{\pars{~x\mbox{-}independent~}}$ !!!.

Felix Marin
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