Brilliant Sub Junior Calculus Contest (Season-1) Note-3

$I=\displaystyle \int { \frac { -({ x }^{ 3 }+3x+1)+6{ x }^{ 3 }+6x }{ { ({ x }^{ 3 }+3x+1) }^{ 3 } } dx }$

This can be further written as,

$I=\displaystyle \int { \frac { -{ ({ x }^{ 3 }+3x+1) }^{ 2 }+{ 2x({ x }^{ 3 }+3x+1) }({ 3x }^{ 2 }+3) }{ { ({ x }^{ 3 }+3x+1) }^{ 4 } } dx }$

This is like the derivative form of $\displaystyle \frac { -x }{ { ({ x }^{ 3 } }+3x+1)^{ 2 }}$

The integral simplifies to,

$I=\displaystyle \frac { -x }{ { ({ x }^{ 3 } }+3x+1)^{ 2 } } +constant$

Now putting the limits $0$ to $1$ we get the value as $-1/25$

Just for enlightening, I'm posting my method too.

$\begin{aligned}\int\frac{5x^3 + 3x - 1}{\left(x^3 + 3x + 1\right)^3}\, dx &=\int\frac{5x^3 + 3x-1}{\left(\sqrt{x}\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)\right)^3}\,dx\\&=\int\frac{5x^3 + 3x-1}{x^{3/2}\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^3}\, dx\\&=\int\frac{5x^{3/2}+3x^{-1/2}-x^{-3/2}}{\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^3}\, dx\\&=\int\frac{2\,d(x^{5/2}+3\sqrt{x}+x^{-1/2})}{\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^3}\\& -\frac{1}{\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^2}+C\\& \end{aligned}$

$\implies \int_{0}^{1} \frac{5x^3 + 3x - 1}{\left(x^3 + 3x + 1\right)^3} dx = \boxed{-\dfrac{1}{25}}$

Can you verify the problem?

I think the solution should be posted

@Saarthak Marathe Why did you change the problem? I solved the original problem and came here to post the solution, only to find that it has been changed.

@Saarthak Marathe Can you post your solution to this question? Thanks.

Using partial fractions,

$\displaystyle \int_{0}^{1} \dfrac{x^4(1-x)^4}{1+x^2} dx = \int_{0}^{1} \left( x^6 -4x^5 +5x^4 -4x^2 - \dfrac{4}{x^2+1} +4 \right) dx$

$= \boxed{\dfrac{22}{7} - \pi}$

Interesting Note : Since the integrand is positive, the integral must also be positive, giving us $\pi < \dfrac{22}{7}$ which is the widely used approximation for $\pi$ .

It is a JEE problem yeah?(This came in the year my sister wrote IIT :P)

Substitute $\cos x = t$, the integral becomes

$\displaystyle \text{I} = \int_{0}^{1} \dfrac{\log t}{1-t^2} \ dt$

$\displaystyle = \sum_{n=0}^{\infty} \int_{0}^{1} t^{2n} \log t \ dt \quad \left(\text{Using infinite G.P.} \right)$

$\displaystyle \int_{0}^{1} t^{2n} \log t \ dt = - \dfrac{1}{(2n+1)^2}$ which can be shown using integration by parts many times.

$\displaystyle \implies \text{I} = - \sum_{n=0}^{\infty} \dfrac{1}{(2n+1)^2}$

$\displaystyle = - \left(\sum_{n=1}^{\infty} \dfrac{1}{n^2} - \sum_{n=1}^{\infty} \dfrac{1}{(2n)^2} \right)$

$\displaystyle = - \dfrac{3}{4} \sum_{n=1}^{\infty} \dfrac{1}{n^2}$

$\displaystyle = \boxed{-\dfrac{\pi^2}{8}}$

Hey,some of my friends told me that it is either quite hard or not possible to solve this without using special functions,does your solution also use any such functions? @Samuel Jones

$\text{Let the integral be J = } \int_{0}^{\frac{\pi}{2}} \frac{log(cosx)}{sinx}dx$

$\text{Put cosx = t }$

$-sinx dx=dt\implies dx = -\frac{dt}{\sqrt{1-t^2}}$

$\text{We have } J = -\int_{0}^{1}\frac{ln(t)}{1-t^2}$

$\text{Put ln(t) = u }\implies \frac{1}{t}dt=du\implies dt=tdu=e^u du$

$J = -\int_{\infty}^{0} \frac{xe^x}{1-e^{2x}}dx = -\frac{1}{2}\int_{\infty}^{0} \frac{x[(e^x+1)+(e^x-1)]}{1-e^{2x}}dx$

$2J = -\int_{\infty}^{0} [\frac{x}{1-e^x} - \frac{x}{e^x+1}]dx$

$2J = -\color{#D61F06}{\underbrace{\int_{0}^{\infty} [\frac{x}{e^x-1}}} + \color{#3D99F6}{\underbrace{\frac{x}{e^x+1}]dx}}$

$\text{In this step we are going to implement some of the special functions , their definitions are as below : }$

$\text{Polygarithm Function :}$

$\large Li_{s}(z) = \frac{1}{\Gamma({s})}\int_{0}^{\infty} \frac{x^{s-1}}{\frac{e^x}{z}-1}dx$

$\large -Li_{s}(-z) = \frac{1}{\Gamma({s})}\int_{0}^{\infty} \frac{x^{s-1}}{\frac{e^x}{z}+1}dx$

$\text{The special values where s=2 \& z=1,-1 are : } Li_{2}(1) = \frac{(\pi)^2}{6} , Li_{2}(-1) = -\frac{(\pi)^2}{12}$

$-2J = \color{#D61F06}{\underbrace{Li_{2}(1)}} + \color{#3D99F6}{\underbrace{-Li_{2}(-1)}}\implies -2J = \frac{(\pi)^2}{6}+\frac{(\pi)^2}{12}\implies \boxed{J=-\frac{\pi^2}{8}}$

$\text{There might be an method to do this by elementary functions . Please post if anyone gets this solved. }$

You should post the solution now.

Take $tanx=t$

Then the integral becomes,

$I=\displaystyle \int { \frac { dt }{ (1+{ t }^{ 2 })(1+{ 2t }^{ 2 }) } }$

Now separate the terms,

$I=\displaystyle 2\int { \frac { dt }{ (1+2{ t }^{ 2 }) } } -\int { \frac { dt }{ ({ t }^{ 2 }+1) } }$

which on integration and resubstituting the value $t= \tan x$ comes out as,

$\sqrt { 2 } { \tan }^{ -1 }(\sqrt { 2 } \tan x)-x+C$

Solution:

$I=\displaystyle \int { \frac { x }{ ({ (2-x)(x-5) })^{ 3/2 } } dx }$

$I= \displaystyle \int { \frac { x }{ \frac { { (x-5) }^{ 3/2 } }{ { (2-x) }^{ 3/2 } } { (2-x) }^{ 3 } } dx }$

Take $t=\frac{x-5}{2-x}$ which gives $x=\frac{2t+5}{1+t}$

$I=\displaystyle \frac { 1 }{ 9 } \int { \frac { \frac { (2t+5) }{ (1+t) } }{ { t }^{ 3/2 }\frac { 1 }{ (1+t) } } } dt$

This simplifies to ,

$I=\displaystyle \frac { 2 }{ 9 } \int { { t }^{ -1/2 }dt+\frac { 5 }{ 9 } \int { { t }^{ -3/2 }dt } }$

This on integration gives,

$I=\displaystyle \frac { 4 }{ 9 } { t }^{ 1/2 }-\frac { 10 }{ 9 } { t }^{ -1/2 }$

Resubstituting $t=\frac{x-5}{2-x}$ we get our answer.

@Akshay Yadav Should I post the solution to this problem ,as no one has posted one yet ?