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Pattern – 7 : Indefinite Integration

February 25, 2021
Category: Pattern – 7 : Indefinite Integration

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⋆ Pattern – 7

$\frac{1}{ax^2+bx+c}$ , $\ \frac{1}{\sqrt{ax^2+bx+c}}$ , $\ \frac{Ax+B}{ax^2+bx+c}$ , $\ \frac{Ax+B}{\sqrt{ax^2+bx+c}}$

Let,

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

$=\ \int\frac{1}{x^2+3x+\frac{9}{4}-\frac{9}{4}+3}\ dx$

$=\ \int\frac{1}{\left(x+\frac{3}{2}\right)^2+\frac{3}{4}}\ dx$

$L.T.\ =\ \frac{\left(M.T.\right)^2}{4\times f.t.}$

$=\ \frac{9x^2}{4\times x^2}$

$=\frac{9}{4}$

$\therefore\ \ I=\ \int\frac{1}{\left(x+\frac{3}{2}\right)^2+\left(\frac{\sqrt3}{2}\right)^2}\ dx$

$=\ \frac{1}{\frac{\sqrt3}{2}}{tan}^{-1}{\left(\frac{x+\frac{3}{2}}{\frac{\sqrt3}{2}}\right)}+c$

$=\ \frac{2}{\sqrt3}{tan}^{-1}{\frac{2x+3}{\sqrt3}}+c$

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

$=\ \int\frac{1}{\sqrt{x^2+3x+\frac{9}{4}-\frac{9}{4}+3}}\ dx$

$=\ \int\frac{1}{\sqrt{\left(x+\frac{3}{2}\right)^2+\frac{3}{4}}}\ dx$

$L.T.\ =\ \frac{\left(M.T.\right)^2}{4\times f.t.}$

$=\ \frac{9x^2}{4\times x^2}$

$=\frac{9}{4}$

$\therefore\ \ I=\ \int\frac{1}{\sqrt{\left(x+\frac{3}{2}\right)^2+\ \left(\frac{\sqrt3}{2}\right)^2}}\ dx$

$=\ \frac{1}{\frac{\sqrt3}{2}}{tan}^{-1}{\left(\frac{x+\frac{3}{2}}{\frac{\sqrt3}{2}}\right)}+c$

$=\ \frac{2}{\sqrt3}{tan}^{-1}{\frac{2x+3}{\sqrt3}}+c$

$\therefore\ \ I=log{\left|x+\frac{3}{2}+\ \sqrt{x^2+3x+3}\right|}$

$\int\frac{1}{\sqrt{x^2+a^2}}\ dx=log{\left|x+\ \sqrt{x^2+a^2\ }\right|}$

$3)\ I=\ \int\frac{3x+5\ }{x^2+5x+11\ }\ dx$

$x^2+5x+11=x^2+5x+\frac{25}{4}-\ \frac{25}{4}$

$=\ \left(x+\frac{5}{2}\right)^2+\frac{19}{4}$

$=\ \left(x+\frac{5}{2}\right)^2+\ \left(\frac{\sqrt{19}}{2}\right)^2---\left(1\right)$

$\therefore\ \ I=\ \int\frac{\frac{3}{2}\ \left(2x+4\right)-\frac{5}{2}}{x^2+5x+11}\ \ dx$

$=\frac{3}{2}\ \int\frac{2x+5}{x^2+5x+11}\ \ dx-\frac{5}{2}\ \int\frac{1}{x^2+5x+11}$

$=\frac{3}{2}log{\left|x^2+5x+11\right|}-\frac{5}{2}\times$ $\frac{1}{\frac{\sqrt{19}}{2}}\ {tan}^{-1}{\frac{x+\frac{3}{2}}{\frac{\sqrt{19}}{2}}}+c$

$=\frac{3}{2}log{\left|x^2+5x+11\right|}$ $\ -\frac{5}{\sqrt{19}}\times{tan}^{-1}{\frac{2x+5}{\sqrt{19}}}+c$

$4)\ I=\ \int\frac{3x+5}{\sqrt{x^2+5x+11}}\ dx$

$x^2+5x+11=x^2+5x+\frac{25}{4}-\frac{25}{4}+11$

$=\ \left(x+\frac{5}{2}\right)^2+\frac{19}{4}$

$=\ \left(x+\frac{5}{2}\right)^2+\ \left(\frac{\sqrt{19}}{2\ }\right)^2$ —-(1)

$\therefore\ \ I=\ \int\frac{\frac{3}{2}\ \left(2x+5\right)-\frac{5}{2}}{\sqrt{x^2+5x+11}}\ \ dx$

$=\frac{3}{2}\ \int\frac{2x+5}{\sqrt{x^2+5x+11}}\ dx$ $-\frac{5}{2}\ \int\frac{1}{x^2+5x+11}\ dx$

$=\frac{3}{2}log{\left|x^2+5x+11\right|}$ $-\frac{5}{2}\times\frac{1}{\frac{\sqrt{19}}{2}}{tan}^{-1}{\frac{x+\frac{5}{2}}{\frac{\sqrt{19}}{2}}}+c$

$=\frac{3}{2}log{\left|x^2+5x+11\right|}$ $-\ \frac{5}{\sqrt{19}}\times{tan}^{-1}{\frac{2x+5}{\sqrt{19}}}+c$

$\therefore\ \ I=\frac{3}{2}\int\frac{2x+5}{\sqrt{x^2+5x+11}}\ dx$ $-\frac{5}{2}\int\frac{1}{\sqrt{x^2+5x\ 11}}\ dx\$

$=\ \frac{-3}{2}\ {\int\left(x^2+5x+11\right)}^\frac{-1}{2}\ \left(2x+5\ \right)\ dx$ $-\frac{5}{2}\int\frac{1}{\sqrt{\left(x+\frac{5}{2}\right)^2+\left(\frac{\sqrt{19}}{2}\right)^2}}\ dx$

$x^2+5x+11=$ $\ \left(x+\frac{5}{2}\right)^2+\ \left(\frac{\sqrt{19\ }}{2}\right)^2---\left(1\right)$

$=\frac{3}{2}\ \frac{\left(x^2+5x+11\right)^\frac{1}{2}}{\frac{1}{2}}-\frac{5}{2}$ $\times l o g{\left|\left(x+\frac{5}{2}\right)+\ \sqrt{x^2+5x+11}\right|}+c$