# What is int_(1)^(4) x^4-x^3+sqrt(x-1)/x^2 dx ?

Dec 19, 2015

$\frac{1023}{5} - \frac{225 - \sqrt{3}}{4} + \arctan \left(\sqrt{3}\right)$

#### Explanation:

This explanation is a bit long, but I couldn't find a quicker way to do it...

The integral is a linear application, so you can already split the function under the integral sign.

${\int}_{1}^{4} \left({x}^{4} - {x}^{3} + \left(\frac{\sqrt{x - 1}}{x} ^ 2\right)\right) \mathrm{dx}$ = ${\int}_{1}^{4} {x}^{4} \mathrm{dx} - {\int}_{1}^{4} {x}^{3} \mathrm{dx} + {\int}_{1}^{4} \frac{\sqrt{x - 1}}{x} ^ 2 \mathrm{dx}$

The 2 first terms are polynomial functions, so they're easy to integrate. I show you how to do it with ${x}^{4}$.

$\int {x}^{4} \mathrm{dx} = {x}^{5} / 5$ so ${\int}_{1}^{4} {x}^{4} \mathrm{dx} = {4}^{5} / 5 - \frac{1}{5} = \frac{1023}{5}$. You do the exact same thing for ${x}^{3}$, the result is $\frac{255}{4}$.

Finding $\int \frac{\sqrt{x - 1}}{x} ^ 2 \mathrm{dx}$ is a bit long and complicated. First you multiply the fraction by $\frac{\sqrt{x - 1}}{\sqrt{x - 1}}$ and then you change the variable : let's say $u = \sqrt{x - 1}$. So $\mathrm{du} = \frac{1}{2 \sqrt{x - 1}} \mathrm{dx}$ and you now have to find $2 \int {u}^{2} / {\left({u}^{2} + 1\right)}^{2} \mathrm{du}$. In order to find it, you need the partial fraction decomposition of the rational function ${x}^{2} / {\left({x}^{2} + 1\right)}^{2}$.

${x}^{2} / {\left({x}^{2} + 1\right)}^{2} = \frac{a x + b}{{x}^{2} + 1} + \frac{c x + d}{{x}^{2} + 1} ^ 2$ with $a , b , c , d \in \mathbb{R}$. After calculus, we find out that ${x}^{2} / {\left({x}^{2} + 1\right)}^{2} = \frac{1}{{x}^{2} + 1} - \frac{1}{{x}^{2} + 1} ^ 2$, which means that $2 \int {u}^{2} / {\left({u}^{2} + 1\right)}^{2} \mathrm{du} = 2 \left(\int \frac{\mathrm{du}}{{u}^{2} + 1} - \int \frac{\mathrm{du}}{{u}^{2} + 1} ^ 2\right)$

$\int \frac{\mathrm{du}}{{u}^{2} + 1} ^ 2$ is well known, it is $\arctan \frac{u}{2} + \frac{u}{2 \left(1 + {u}^{2}\right)}$.

Finally, $2 \int {u}^{2} / {\left({u}^{2} + 1\right)}^{2} \mathrm{du} = 2 \left(\arctan \left(u\right) - \arctan \frac{u}{2} - \frac{u}{2 \left(1 + {u}^{2}\right)}\right) = \arctan \left(u\right) - \frac{u}{1 + {u}^{2}}$

You replace $u$ by its original expression with $x$ to have $\int \frac{\sqrt{x - 1}}{x} ^ 2 \mathrm{dx}$, which is $\arctan \left(\sqrt{x - 1}\right) - \frac{\sqrt{x - 1}}{x}$

So finally, ${\int}_{1}^{4} \frac{\sqrt{x - 1}}{x} ^ 2 \mathrm{dx} = \arctan \left(\sqrt{3}\right) - \frac{\sqrt{3}}{4}$