Question

Find the vaue of
(I)$( \int\frac{3x+5}{x^2+4x+7}dx$
(II) $\int\limits {\frac{2x+3}{\sqrt{x^2-3x+1}}} \, dx$

Answer 1

$\text{(I)} \int {\frac{3x+5}{x^2-3x+1}}dx$

$I = \int {\frac{3x+5}{x^2-3x+1}}$

$3x+5 = A \frac{d}{dx}(x^2+4x+7)+B$

$3x+5 = A(2x+4)+B$

$2A = 3 \implies A = \frac{3}{2}\:\:;4A+B = 5 \implies B = -1$

$I = \int \frac{\frac{3}{2}(2x+4)-1}{x^2+4x+7}dx$

$I = \frac{3}{2} \int \frac{2x+4}{x^2+4x+7}dx - \int \frac{1}{x^2+4x+7}dx$

$= \frac{3}{2} \log |x^2+4x+7|+c - \int \frac{1}{(x+2)^2 + (\sqrt{3})^2} dx$

$\boxed {I=\frac{3}{2}\log|x^2+4x+7|-\frac{1}{3}\tan^{-1} (\frac{x+2}{\sqrt{3}})+c}$

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$\text{(II)} \int \frac{2x+3}{\sqrt{x^2+x+1}}dx$

$I = \int \frac{2x+3}{\sqrt{x^2+x+1}}dx$

$2x+3 = A \frac{d}{dx}(x^2+x+1)+B$

$2x+3 = A(2x+1)+B$

$2A = 2 \implies A = 1 \: : \: A+B = 3 \implies B = 2$

$I = \int \frac{(2x+1)+2}{\sqrt{x^2+x+1}}dx$

$I = \int \frac{2x+1}{\sqrt{x^2+x+1}}dx+2\int\frac{1}{\sqrt{x^2+x+1}}dx$

$= 2\sqrt{x^2+x+1}+2\int\frac{1}{\sqrt{(x + \frac{1}{2})^2+ (\frac{\sqrt{3}}{2})^2}}dx$

$= 2\sqrt{x^2+x+1}+2\log|x+\frac{1}{2}+\sqrt{(x+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}+c$

$\boxed{I = 2 \sqrt{x^2+x+1}+2\log|x+\frac{1}{2}+\sqrt{x^2+x+1}|+c}$

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${\text{\large{\underline{\underline{Some Integral Values :}}}}}$

$\int \frac{dx}{a^2-x^2} = \frac{1}{2a}\log|\frac{a+x}{a-x}|+c$

$\int \frac{dx}{x^2-a^2} = \frac{1}{2a}\log|\frac{x-a}{x+a}|+c$

$\int \frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}(\frac{x}{a})+c$

$\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(\frac{x}{a})+c$

$\int \frac{dx}{\sqrt{x^2-a^2}} = \log|x+\sqrt{x^2-a^2}|+c$

$\int \frac{dx}{x^2+a^2} = \log|x+\sqrt{x^2+a^2}|+c$