# Cant integrate

Note by Hemant Khatri
3 years, 10 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:

I tell you question put up by you is not too hard but way too lengthy. Anyway here's the solution.

$$I=\int { \frac { dx }{ { x }^{ 3 }+{ x }^{ 8 } } } =\int { \frac { dx }{ { x }^{ 3 }(1+{ x }^{ 5 }) } } =\int { \frac { dx }{ { x }^{ 3 } } } -\int { \frac { { x }^{ 2 }dx }{ 1+{ x }^{ 5 } } }$$

Now we can write : $${x}^{5}+1=(x+1)({x}^{2}+2xcos(\frac{2\pi}{5})+1)({x}^{2}-2xcos(\frac{\pi}{5})+1)$$

Hence our $$I$$ becomes :

$$I=\frac{-2}{{x}^{2}}-{I}_{1}$$

$${I}_{1}=\int { \frac { { x }^{ 2 }dx }{(x+1)({x}^{2}+2xcos(\frac{2\pi}{5})+1)({x}^{2}-2xcos(\frac{\pi}{5})+1) } }$$

after some tedious applications of partial fractions we get :

$${ I }_{ 1 }=\frac { 1 }{ 5 } (\frac { 1 }{ x+1 } +(\frac { 1}{ 2cos(\frac { \pi }{ 5 } ) } )(\frac { x+1 }{ { x }^{ 2 }-2xcos(\frac { \pi }{ 5 } )+1 } )-(\frac { 1 }{ 2cos(\frac { 2\pi }{ 5 } ) } )(\frac { x+1 }{ { x }^{ 2 }+2xcos\frac { 2\pi }{ 5 } +1 } ))$$

$$\Rightarrow { I }_{ 1 }=\frac { 1 }{ 5 } log(x+1)+\frac { 1 }{ 10cos(\frac { \pi }{ 5 } ) } { J }_{ 1 }-\frac { 1 }{ 10cos(\frac { 2\pi }{ 5 } ) } { J }_{ 2 }$$

$${ J }_{ 1 }=\int { \frac { (x+1)dx }{ ({ x }^{ 2 }-2xcos\frac { \pi }{ 5 } +1) } } =\frac { 1 }{ 2 } (\int { \frac { 2x-2cos\frac { \pi }{ 5 } }{ { x }^{ 2 }-2xcos\frac { \pi }{ 5 } +1 } dx } )+(1+cos\frac { \pi }{ 5 } )\int { \frac { dx }{ { ( x -cos\frac { \pi }{ 5 } ) }^{ 2 }+{ (sin\frac { \pi }{ 5 } })^{ 2 } } }$$

In the expression for $${J}_{1}$$ in the first integral put $$({ x }^{ 2 }-2xcos\frac { \pi }{ 5 } +1)=t$$ and solve and the second integral can be solved putting $$(x -cos\frac { \pi }{ 5 } )=tan(t)sin(\frac{\pi}{5})$$ and solve to get :

$${J}_{1}=\frac{1}{2}ln({ x }^{ 2 }-2xcos\frac { \pi }{ 5 } +1)+(\frac { 1+cos\frac { \pi }{ 5 } }{ sin\frac { \pi }{ 5 } } ){ tan }^{ -1 }(\frac { x-cos\frac { \pi }{ 5 } }{ sin\frac { \pi }{ 5 } } )$$

Similarly $${ J }_{ 2 }=\int { \frac { (x+1)dx }{ ({ x }^{ 2 }+2xcos\frac { 2\pi }{ 5 } +1) } } =\frac { 1 }{ 2 } (\int { \frac { 2x+2cos\frac {2 \pi }{ 5 } }{ { x }^{ 2 }+2xcos\frac {2\pi }{ 5 } +1 } dx } )+(1-cos\frac {2\pi }{ 5 } )\int { \frac { dx }{ { ( x +cos\frac { 2\pi }{ 5 } ) }^{ 2 }+{ (sin\frac { 2\pi }{ 5 } })^{ 2 } } }$$

Solving it similarly we get :

$${J}_{2}=\frac{1}{2}ln({ x }^{ 2 }+2xcos\frac {2\pi }{ 5 } +1)+(\frac { 1-cos\frac { 2\pi }{ 5 } }{ sin\frac { 2\pi }{ 5 } } ){ tan }^{ -1 }(\frac { x+cos\frac { 2\pi }{ 5 } }{ sin\frac { 2\pi }{ 5 } } )$$

Putting back all the values we get I as :

$$I=\frac { -2 }{ { x }^{ 2 } } -\frac { 1 }{ 5 } log(x+1)-\frac { 1 }{ 20cos\frac { \pi }{ 5 } } ln({ x }^{ 2 }-2xcos\frac { \pi }{ 5 } +1)+\frac { 1 }{ 20cos\frac { 2\pi }{ 5 } } ln({ x }^{ 2 }+2xcos\frac { 2\pi }{ 5 } +1)-\frac { 1+cos\frac { \pi }{ 5 } }{ 10sin\frac { \pi }{ 5 } cos\frac { \pi }{ 5 } } { tan }^{ -1 }(\frac { x-cos\frac { \pi }{ 5 } }{ sin\frac { \pi }{ 5 } } )\\+\frac { 1-cos\frac { 2\pi }{ 5 } }{ 10sin\frac { 2\pi }{ 5 } cos\frac { 2\pi }{ 5 } } { tan }^{ -1 }(\frac { x+cos\frac { 2\pi }{ 5 } }{ sin\frac { 2\pi }{ 5 } } )$$

- 3 years, 10 months ago

Awesome Solution...But can't it get shorter, bro?

- 3 years, 10 months ago

There's no way to get it shorter.

- 3 years, 10 months ago

Very much complicated bro :D Anyways nice solution

- 3 years, 10 months ago

Such integrals are always complicated.

- 3 years, 10 months ago

Yeah u r rite

- 3 years, 10 months ago

Yeah i agreed... Please solve this

- 3 years, 10 months ago