Solve an indefinite integral

Question

Find the value of $$ \int\frac{\sqrt{1+x^2}}{1-x^2}dx $$

My Attempt: I tried to arrange the numerator as follows, $$ \sqrt{1+x^2} = \sqrt{1-x^2 + 2x^2} $$ but that didn't help me. Any help will be appreciated.

Answer 1

HINT: Make the substitution $1+x^2 \to u$ or $x \to \sqrt{u-1}$. Then you can transform the integral into $$\int -\frac{\sqrt u}{u}\cdot\frac{1}{2\sqrt{u-1}}du$$ $$\int -\frac{1}{\sqrt u}\cdot\frac{1}{2\sqrt{u-1}}du$$ $$-\frac{1}{2}\int \frac{1}{\sqrt{u^2-u}}du$$ Then complete the square in the denominator: $$-\frac{1}{2}\int \frac{1}{\sqrt{(u-\frac{1}{2})^2-\frac{1}{4}}}du$$ Now see if you can get it in the form $$\int \frac{1}{\sqrt{a^2-1}}$$ so that you can use the formula $$\int \frac{1}{\sqrt{a^2-1}}=\ln\bigg(\sqrt{x^2-1}+x\bigg)+C$$

Answer 2

Try $x=\cos \phi$. Then you get $1+x^2=2\cos^2 \frac{\phi}{2}$, $1-x^2=\sin^2\phi$ and $dx=-\sin\phi\,d\phi$. Finally you will use $\sin\phi=2\sin\frac{\phi}{2}\cos\frac{\phi}{2}$ and you get an integral you can find in most tables.

Answer 3

Hint 1: Using $x=\tan(u)$, we get $$ \begin{align} \int\frac{\sqrt{1+x^2}}{1-x^2}\,\mathrm{d}x &=\int\frac{\sec^3(u)}{1-\tan^2(u)}\,\mathrm{d}u\\ &=\int\frac{\sec(u)}{\cos^2(u)-\sin^2(u)}\,\mathrm{d}u\\ &=\int\frac{\sec^2(u)}{1-2\sin^2(u)}\,\mathrm{d}\sin(u)\\ &=\int\frac{\mathrm{d}\sin(u)}{\left(1-2\sin^2(u)\right)\left(1-\sin^2(u)\right)}\\ \end{align} $$ Hint 2: Use partial fractions: $$ \begin{align} &\frac1{\left(1-2v^2\right)\left(1-v^2\right)}\\ &=\frac2{1-2v^2}-\frac1{1-v^2}\\ &=\frac1{1-\sqrt2v}+\frac1{1+\sqrt2v}-\frac{1/2}{1-v}-\frac{1/2}{1+v} \end{align} $$

Answer 4

I dont know complex integration but you can substitute $x=\tan (t) $ we get $\int \frac {\sec^3 (t)}{1-\tan^2 (t)} $ which simplifies to $\int \frac {1}{\cos (x)\cos (2x)} $ from here write it as $\int \frac {1}{(\cos (x))(\cos^2 (x)-\sin^2 (x))} $ which simplies to $\frac {\cos (x)}{(1-sin^2 (x))(1-2sin^2 (x)} $ use $\sin (x)=u $ hence integral changes to $\int \frac {1}{(1-u^2)(1-2u^2)} $ which can be easily done using partial fractions. And the integral can be calculated quite easily. Hope you are fine with real integration.