Find $$\int \frac{x^2}{(5+x^2)^2} \, dx.$$ Find $$\int \frac{dx}{\sqrt{4x^2+1}}.$$ Evaluate $$\int_0^1 \tan^{-1}(x) \, dx.$$ Evaluate $$\int_2^2 \frac{dx}{\sqrt{2x-1}}.$$ It can be shown that $$\frac{8(1-x)}{(2-x^2)(2-2x+x^2)} = \frac{4-2x}{2-2x+x^2}-\frac{2x}{2-x^2} .\ (Do NOT prove this.)$$ Use this result to evaluate $$\int_0^1 \frac{8(1-x)}{(2-x^2)(2-2x+x^2)} \, dx.$$ - HSC - SSCE Mathematics Extension 2 - Question 1 - 2008 - Paper 1

Using the substitution ( x = \frac{1}{2} \tan(\theta) ), we get ( dx = \frac{1}{2} \sec^2(\theta) , d\theta ). Then:

$$\int \frac{dx}{\sqrt{4x^2+1}} = \int \frac{\frac{1}{2} \sec^2(\theta)}{\sqrt{1+\tan^2(\theta)}} , d\theta = \int \sec(\theta) , d\theta = \ln |\sec(\theta) + \tan(\theta)| + C = \ln |\sqrt{4x^2 + 1} + 2x| + C.$

We can use integration by parts, letting ( u = \tan^{-1}(x) ) and ( dv = dx ). Then, ( du = \frac{1}{1+x^2} , dx ) and ( v = x ). Thus:

$\int_0^1 \tan^{-1}(x) \, dx = \left[ x \tan^{-1}(x) \right]_0^1 - \int_0^1 \frac{x}{1+x^2} \, dx = \frac{\pi}{4} - \left[ \frac{1}{2} \ln(1+x^2) \right]_0^1 = \frac{\pi}{4} - \frac{1}{2}\ln(2).$

Here, the integral simplifies as the limits of integration are equal. Therefore,

$\int_2^2 \frac{dx}{\sqrt{2x-1}} = 0.$

Using the identity provided:

$\int_0^1 \frac{8(1-x)}{(2-x^2)(2-2x+x^2)} \, dx = \int_0^1 \left( \frac{4-2x}{2-2x+x^2} - \frac{2x}{2-x^2} \right) \, dx.$

We can then integrate each part separately, which would involve resolving each rational function using partial fractions.