# Simple but Difficult Integral?

Hello brilliant users, I'm Leonardo. I've some trouble to evaluate this integral. It took me several days to realize that this problem need some extraordinary way to solve it. Can somebody show me how to get the answer? Thanks

Note by Leonardo Chandra
4 years, 11 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Substituting $$x=u^{2}$$ and we have, $$dx = 2udu$$ Our integral becomes :

$\Rightarrow \int \frac{\sqrt{x}}{1+x^{2}} dx = \int \frac{2u^{2}}{1+u^{4}} du$

Adding and subtracting $$1$$ in numerator,

$\Rightarrow \int \frac{2u^{2}}{1+u^{4}}du$

$\Rightarrow \int \frac{(u^{2}+1)+(u^{2}-1)}{1+u^{4}}du$

$\Rightarrow \int \frac{u^{2}+1}{u^{4}+1}du + \int \frac{u^{2}-1}{u^{4}+1}du$

Dividing by $$u^{2}$$ in numerator and denominator,

$\Rightarrow \int \frac{1+\frac{1}{u^{2}}}{u^{2}+ \frac{1}{u^{2}}}du + \int \frac{1-\frac{1}{u^{2}}}{u^{2}+ \frac{1}{u^{2}}}du$

Making perfect square in denominator,

$\Rightarrow \int \frac{1+\frac{1}{u^{2}}}{(u-\frac{1}{u})^{2}+2}du + \int \frac{1-\frac{1}{u^{2}}}{(u+\frac{1}{u})^{2}-2}du$

Substituting $$u-\frac{1}{u} = t$$ and $$u+\frac{1}{u} = v$$ we get, $$(1+\frac{1}{u^{2}})du = dt$$ and similarly, $$(1-\frac{1}{u^{2}})du = dv$$

Therefore, our integral becomes,

$\Rightarrow \int \frac{dt}{t^{2}+(\sqrt{2})^{2}} + \int \frac{dv}{v^{2}-(\sqrt{2})^{2}}$

$\Rightarrow \frac{1}{\sqrt{2}} tan^{-1}(\frac{t}{\sqrt{2}}) + \frac{1}{2\sqrt{2}}\log _e \left( \frac{v-\sqrt{2}}{v+\sqrt{2}} \right) + C$

Putting values of t and v, we get

$\Rightarrow \frac{1}{\sqrt{2}} tan^{-1}(\frac{u-\frac{1}{u}}{\sqrt{2}}) + \frac{1}{2\sqrt{2}}\log _e \left( \frac{u+\frac{1}{u}-\sqrt{2}}{u+\frac{1}{u}+\sqrt{2}} \right) + C$

Now, putting value of u, we get

$\Rightarrow \frac{1}{\sqrt{2}} tan^{-1}(\frac{\sqrt{x}-\frac{1}{\sqrt{x}}}{\sqrt{2}}) + \frac{1}{2\sqrt{2}}\log _e \left( \frac{\sqrt{x}+\frac{1}{\sqrt{x}}-\sqrt{2}}{\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{2}} \right) + C$

- 4 years, 11 months ago

The key is to use a substitution, e.g. $$u=\sqrt{x}$$ which gives $$du=\dfrac1{2\sqrt{x}}\,dx\Leftrightarrow 2u\,du=dx$$. Observe our integral becomes:$\int\frac{\sqrt{x}}{1+x^2}dx=2\int\frac{u^2}{1+u^4}\,du$next, factor using Germain's identity and decompose into partial fractions.

- 4 years, 11 months ago