# Problem on indefinite integration

$$f(x) = \int\frac{x^2}{(1+x^2)(1 + \sqrt{1+x^2})}dx$$ where $f(0)=0$, find $f(1)$.

Despite many tries by taking $$\sqrt{1+x^2}=t$$ or $$1+\sqrt{1+x^2}$$ as $t$, I could not get it solved.

HINT:

Let $x=\tan t$

$$\int\dfrac{x^2}{(1+x^2)({1+\sqrt{1+x^2)}}}dx=\int\dfrac{\tan^2t dt}{1+\sec t}$$

$$=\int\dfrac{1-\cos^2t}{\cos t(1+\cos t)}=\int\dfrac{1-\cos t}{\cos t}dt$$

• Thanks. I forgot trigonometric substitutions,,, Feb 25, 2017 at 13:06

Hint:

$$I = \int \frac{x^2}{(1+x^2)(1+\sqrt{1+x^2})} \mathrm{d}x = \frac{x^2(\sqrt{1+x^2}-1)}{(1+x^2)(1+\sqrt{1+x^2})(\sqrt{1+x^2}-1)} \mathrm{d}x =\int \frac{\sqrt{x^2+1}-1}{1+x^2}\mathrm{d}x = \int \frac{1}{\sqrt{1+x^2}}-\frac{1}{1+x^2}\mathrm{d}x$$

Hope you can take it from here.

• Thanks rationalisation also works ,,, Feb 25, 2017 at 13:12

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

With Euler Substitution

$$x = {1 - t^{2} \over 2t}\,\qquad t = \root{1 + x^{2}} - x$$ $\ds{\mrm{f}\pars{x}}$ becomes \begin{align} \mrm{f}\pars{x} & = \int\pars{-\,{1 \over t} + {2 \over t^{2} + 1}}\,\dd t = -\ln\pars{t} + 2\arctan\pars{t} \\[5mm] & = 2\arctan\pars{\root{1 + x^{2}} - x} - \ln\pars{\root{1 + x^{2}} - x} + \pars{~\mbox{a constant}~} \end{align}