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Pattern - 38 of Indefinite Integration

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Pattern – 38 :

$\int\frac{1}{\left(px+q\right)\sqrt{ax+b}}dx,$ then put ax + b + $t^2$

1) Evaluate : $\int\frac{1}{\left(2x+3\ \right)\sqrt{4x-7}}dx$

Solution : Let

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

Let, 4x – 7 = $t^2$

∴ 4x = $t^2$ + 7

$\therefore\ \ x=\ \frac{t^2+7}{4}$

4dx = 2t dt

$\therefore\ \ dx=\ \frac{tdt}{2}$

$=\ \int\frac{1}{\left[2\left(\frac{t^2+7}{4}\right)+3\right]\ \sqrt{t^2}}\times\frac{t}{2}dt$

$=\frac{1}{2}\ \int\frac{2}{\left(t^2+13\right)}\ dt$

$=\ \int \frac{1}{t^2+13}dt$

$=\ \int\frac{1}{t^2+\ \left(\sqrt{13}\right)^2}\ dt$

$=\ \frac{1}{\sqrt{13}}\tan^{-1}{\frac{t}{\sqrt{13}}}+c$

$=\ \frac{1}{\sqrt{13}}\tan^{-1}{\sqrt{\frac{4x-7}{13}}}+c$