How to integrate $\int \frac{dx}{x^2 \sqrt{x-1}}$?

Question

I need to integrate$$\int \dfrac{dx}{x^2 \sqrt{x-1}}.$$ I've tried everything from substitutions ($\sqrt{x-1}=u$) to integration by parts but I cannot get anywhere. Please help.

Answer 1

The change of variable $$ \sqrt{x-1}=u\text{ or }x=u^2+1,\quad dx=2\,du $$ transforms the integral into $$ 2\int\frac{du}{(u^2+1)^2}. $$ Do you know how to do tjis one?

Answer 2

$$I=\int \dfrac{dx}{x^2 \sqrt{x-1}}$$ Let $x=\sec^2 t,dx=2\sec^2 t\tan t\, dt$: $$ I=\int\dfrac{2\sec^2 t\tan t dt}{\sec^4t\tan t}=\int 2\cos^2tdt=\int(1+\cos 2t)dt=t+\dfrac12\sin2t+c=\arccos\dfrac1{\sqrt x}+\dfrac{\sqrt{x-1}}{x}+c$$

Answer 3

Setting $u=\sqrt{x-1}$ gives you $$ \int \frac{2u \, du}{(1+u^2)^2u}= 2\int \frac{du}{(1+u^2)^2} \tag{1}$$ There are two ways to do this: the first is to change variables again to $u=\tan{\theta}$, so $du=\sec^2{\theta} \, d\theta$: $$ \begin{align*} 2\int \frac{\sec^2{\theta} \, d\theta}{(1+\tan^2{\theta})^2} &= 2\int \frac{\sec^2{\theta} \, d\theta}{(1+\tan^2{\theta})^2} \\ &= \int 2\cos^2{\theta} \, d\theta \\ &= \int (1+\cos{2\theta}) d\theta, \end{align*}$$ using the double-angle formula $\cos{2\theta}=2\cos^2{\theta}-1$. $$ \int (1+\cos{2\theta}) = \theta + \frac{1}{2}\sin{2\theta}, $$ and substituting back, $\theta = \arctan{\sqrt{x-1}}$, and in particular, $$ \int \frac{dx}{x^2\sqrt{x-1}} = \frac{1}{2} \sin{(2\arctan{\sqrt{x-1}})} + \arctan{\sqrt{x-1}}. $$ The first term can be simplified using the formula $\sin{(2 \arctan{t})} = \frac{2t}{1+t^2}$ to give the answer.

Alternatively:

Having got to (1), we can write $1=(1+u^2)-u^2$ integrate by parts: $$ 2\int \frac{(1+u^2)-u^2}{(1+u^2)^2}\,du = \int \frac{2du}{1+u^2} - \int u\frac{2u}{(1+u^2)^2} \, du \\ = \int \frac{2du}{1+u^2} + \frac{u}{1+u^2} - \int \frac{du}{1+u^2} \, du, $$ and then proceed to substitute $u=\tan{\theta}$ in the first one to get the $\arctan$ term. This way, we avoided having to use lots of double angle formulae.

Answer 4

\begin{align} \int \dfrac{1}{x^2 \sqrt{x-1}}dx=&\int \frac{1}{\sqrt{x-1}}d\left(\frac{x-1}x\right)\\ = &\ \frac{\sqrt{x-1}}x +\int \frac{ d(\sqrt{x-1})}{x} = \frac{\sqrt{x-1}}x +\tan^{-1} \sqrt{x-1}+C \end{align}

Answer 5

Since the derivative of $\frac1{x}$ is present in the integrand, the substitution $x=\frac1{t}$ is feasible. So, the integral becomes

$$\mathcal I=-\int\frac{\sqrt t\mathrm dt}{\sqrt{1-t}}$$

Now, notice that in this integral, the derivative of $\sqrt{1-t}$ is present, so we can let $u=\sqrt{1-t}$:

$$\begin{align}\mathcal I&=2\int\sqrt{1-u^2}\mathrm du\\&=u\sqrt{1-u^2}+\sin^{-1}u+C\end{align}$$

Now, just do the back substitution.

Answer 6

Substitute $u=\sqrt{x-1}$. Then $2du=\dfrac{1}{\sqrt{x-1}}dx$ and $x=u^2+1$.

Therefore $$\int \dfrac{1}{x^2 \sqrt{x-1}}dx=\int \dfrac{2}{(u^2+1)^2}du.$$

Put $U=\dfrac{1}{u^{2}+1}$ and $dV=du$. Then integration by parts,

$$\int \dfrac{1}{(u^2+1)^2}du=\dfrac{u}{u^{2}+1}+\int \dfrac{2u^2}{(u^2+1)^2}du\\=\dfrac{u}{u^{2}+1}+2\int \left( \dfrac{u^{2}+1}{(u^{2}+1)^2}-\dfrac{1}{(u^{2}+1)^2}\right)du\\ =\dfrac{u}{u^{2}+1}+2\int \left( \dfrac{1}{(u^{2}+1)}-\dfrac{1}{(u^{2}+1)^2}\right)du $$

Hence $$\int \dfrac{1}{(u^2+1)^2}du=\dfrac{u}{2(u^{2}+1)}+\dfrac{1}{2}\int \dfrac{1}{(u^{2}+1)}du=\dfrac{u}{2(u^{2}+1)}+\dfrac{1}{2}\tan^{-1}(u).$$

Therefore $$\int \dfrac{1}{x^2 \sqrt{x-1}}dx=\dfrac{\sqrt{x-1}}{x}+\tan^{-1}(\sqrt{x-1}).$$