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

For this integral, we can use a trigonometric substitution. Let ( x = \frac{1}{2} \tan \theta ). Then, ( dx = \frac{1}{2} \sec^2 \theta , d\theta ) and we find:

$\int \frac{\frac{1}{2} \sec^2 \theta}{\sqrt{4(\frac{1}{4}\tan^2 \theta + 1)}} \;d\theta = \int \frac{\frac{1}{2} \sec^2 \theta}{\sqrt{\sec^2 \theta}} \;d\theta = \int \frac{1}{2} \;d\theta$

The integral simplifies to ( \frac{1}{2} \theta + C = \frac{1}{2} \tan^{-1}(2x) + C).

To evaluate this integral, we can use integration by parts. Let ( u = \tan^{-1} x ) and ( dv = dx ). Then, we have:

$du = \frac{1}{1+x^2} \;dx$ $v = x$

Thus,

$\int \tan^{-1} x \;dx = x \tan^{-1} x - \int \frac{x}{1+x^2} \;dx$

To evaluate ( \int \frac{x}{1+x^2} ;dx), use the substitution ( w = 1 + x^2 ):

$\int \frac{x}{1+x^2} \;dx = \frac{1}{2}\ln(1+x^2) + C$

Substituting back, we find:

Evaluate from 0 to 1 to find the definite integral value.

Evaluate this definite integral, first observe that if the upper limit equals the lower limit (2), the result will be 0. Therefore:

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

We start by substituting the known identity:

The integral simplifies to:

$\int_0^1 \frac{8(1-x)}{(2-x^2)(2-2x+x^2)} \;dx$,

which requires using the previous evaluation results.

Evaluate it accordingly to find the final specific value.