A lvl H2 Maths: Integration

Question from http://www.sgforums.com/forums/2297/topics/339990

(a) Evaluate
$\int \frac{1+2x}{1+x^{2}} dx$

(b) Using the substitution $x= \sin ^{2}u$ , find
$\int \sqrt{\frac{x}{1-x} }dx$
with x in the range 0 < x < 1.

(c) Find $\int \frac{2x-1}{x^{2}-2x+5}dx$ .

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Answer:

(a)
$\int \frac{1+2x}{1+x^{2}} dx$
$=\int \frac{1}{1+x^{2}} + \frac{2x}{1+x^{2}} dx$
$=\int \frac{1}{1+x^{2}}dx + \int\frac{2x}{1+x^{2}} dx$
$= \tan^{-1} x+ ln\left(1+x^{2} \right)+c$
Note: For $\int\frac{2x}{1+x^{2}} dx$ , the numerator is a differential of the denominator.

(b)
From $x= \sin ^{2}u$ ,
$\frac{dx}{du}=2\sin u \cos u$
$dx=2\sin u \cos u du$
Sub $x= \sin ^{2}u$ and $dx=2\sin u \cos u du$
$\int \sqrt{\frac{\sin ^{2}u}{1-\sin ^{2}u} }\left(2\sin u \cos u \right)du$
$=\int \sqrt{\frac{\sin ^{2}u}{\cos ^{2}u} }\left(2\sin u \cos u \right)du$
$=\int \frac{\sin u}{\cos u} \left(2\sin u \cos u \right)du$
$=\int \left(2\sin^{2} u \right)du$
$=\int \left(1-\cos 2u \right)du$
$=u-\frac{1}{2}\sin 2u + c$
Since $x= \sin ^{2}u$ , $u=\sin^{-1} \sqrt{x}$
Also, $\cos^{2} u = 1 - \sin^{2} u$
$\cos u = \sqrt{1 - \sin^{2} u}$
$\cos u = \sqrt{1 - x}$

Thus, sin 2u = 2 sin u cos u
$\sin 2u = 2\sqrt{x}\sqrt{1-x}$
$\sin 2u = 2\sqrt{x-x^{2}}$

Hence,
$\int \sqrt{\frac{x}{1-x} }dx$ $=\sin^{-1} \sqrt{x}-\sqrt{x-x^{2}}+c$

(c)
$\int \frac{2x-1}{x^{2}-2x+5}dx$
$\small =\int \frac{2x-2}{x^{2}-2x+5}+\frac{1}{x^{2}-2x+5}dx$
$\small =\int \frac{2x-2}{x^{2}-2x+5}dx+\int \frac{1}{x^{2}-2x+5}dx$
$\small =\int \frac{2x-2}{x^{2}-2x+5}dx+\int \frac{1}{\left(x-1\right)^{2}+4}dx$
$\small =\int \frac{2x-2}{x^{2}-2x+5}dx+\frac{1}{4}\int \frac{1}{\left(\frac{x-1}{2}\right)^{2}+1}dx$
$\small =\ln \left( x^{2}-2x+5\right)+\frac{1}{4}\tan^{-1}\left(\frac{x-1}{2}\right)\bullet 2+c$
$\small =\ln \left( x^{2}-2x+5\right)+\frac{1}{2}\tan^{-1}\left(\frac{x-1}{2}\right)+c$

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A lvl H2 Maths: Integration