Archive for October, 2014

a solution to another integral

Posted: October 8, 2014 in Mathematics

This week’s integral:

\displaystyle \int_0^1 \frac{\log(x+1)}{x^2+1} \;dx


At first glance the integral looks scary because there is no way of making the numerator being the derivative of anything in the denominator. So what can we do about it? How about trying to simplify the denominator or get rid of it completely. Let’s make use of the trig sub,

x=\tan\theta, \quad dx=\sec^2{\theta}\;d\theta, \quad \tan^2\theta+1=\sec^2\theta

then we have,

\displaystyle \int_0^{\pi/4}\log(\tan\theta+1)\;d\theta

Now apply the trick of

\displaystyle \int_0^a f(t) \;dt = \int_0^a f(a-t) \;dt

we have,

I=\displaystyle \int_0^{\pi/4}\log\left(\tan\theta+1\right)\;d\theta = \int_0^{\pi/4}\log\left(\tan\left(\frac{\pi}{4}-\theta\right)+1\right)\;d\theta

\displaystyle I = \int_0^{\pi/4}\log\left(\frac{1-\tan\theta}{1+\tan\theta}+1\right)\;d\theta = \int_0^{\pi/4}\log\frac{2}{1+\tan\theta}\;d\theta

\displaystyle I = \int_0^{\pi/4} \log 2 \;d\theta - \int_0^{\pi/4}\log(\tan\theta+1)\;d\theta

\displaystyle \therefore\;I = \frac{\log 2}{2}\cdot\frac{\pi}{4} = \frac{\pi\log2}{8}


a solution to an integral

Posted: October 3, 2014 in Mathematics

\displaystyle\int\frac{1}{x^4+x^2+1}\;\text{d}x, \quad \text{for } x\in\mathbb{R}


Realizing that there is no way we can use a normal “U sub” method we learned from highschool, we need to be a bit creative and find a way to tackle this integral by playing around with the polynomial in the denominator.
First divide top and bottom by x^2,

\displaystyle\int\frac{\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}\text{ d}x


\displaystyle \text{d}\left(x-\frac{1}{x}\right)=1+\frac{1}{x^2}\text{ d}x \quad \text{and} \quad \text{d}\left(x+\frac{1}{x}\right)=1-\frac{1}{x^2}\text{ d}x

We need to somehow make the derivative of the denominator “appears” in the numerator by completing the square in the denominator and add 1 and subtract 1 in both integrals.

\displaystyle \frac{1}{2}\int\frac{1+\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}\text{ d}x - \frac{1}{2}\int\frac{1-\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}\text{ d}x

\displaystyle \frac{1}{2}\int\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+3}\text{ d}x - \frac{1}{2}\int\frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})^2-1}\text{ d}x

Let u=x-\dfrac{1}{x}, \; v=x+\dfrac{1}{x}, we have

\displaystyle \frac{1}{2}\int\frac{\text{d}u}{u^2+3} - \frac{1}{2}\int\frac{\text{d}v}{v^2-1}

Then the rest is routine and easy!

\displaystyle \frac{1}{6}\int\frac{\text{d}u}{(\frac{u}{\sqrt3})^2+1} - \frac{1}{4}\int\frac{1}{v-1}-\frac{1}{v+1}\text{ d}v

\displaystyle \frac{\sqrt3}{6}\tan^{-1}\frac{u}{\sqrt3}-\frac{1}{4}\log\left|\frac{v-1}{v+1}\right|+C

Sub back the x,

\displaystyle \frac{\sqrt3}{6}\tan^{-1}\frac{1}{\sqrt3}\left(x-\frac{1}{x}\right) - \frac{1}{4}\log\left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|+C

We can drop the absolute value sign since x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0 and x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0


\displaystyle \int\frac{1}{x^4+x^2+1}\;\text{d}x = \frac{\sqrt3}{6}\tan^{-1}\left(\frac{x^2-1}{\sqrt3x}\right) - \frac{1}{4}\log\left(\frac{x^2-x+1}{x^2+x+1}\right)+C