Blog

Pattern - 41 of Indefinite Integration

No Comments

⋆ Pattern : 41

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

1) Evaluate : $\int\frac{1}{\left(2x^2+5\right)\sqrt{7x^2+3}}\ dx$

Let, x = $\frac{1}{t}$

∴ t = $\frac{1}{x}$

dx = $\frac{-1}{t^2}dt$

$\left(2x^2+5\right)\sqrt{7x^2+3}$

$=\ \left(\frac{2}{t^2}+5\right)\sqrt{\frac{7}{t^2}+3}$

$=\ \left(\frac{2+5t^2}{t^2}\right)\sqrt{\frac{7+3t^2}{t^2}}$

$=\ \frac{\left(2+5t^2\right)}{t^2}\ \times\frac{\sqrt{7+3t^2}}{t}$

$\therefore\ \ I=\ -\int\frac{t^2\times t}{\left(2+5t^2\right)\sqrt{7+3t^2}}\times\frac{1}{t^2}\ dt$

Let, 7 + 3 $t^2 = z^2$

∴ 3 $t^2 = z^2$ – 7

∴ $t^2 = \frac{z^2-7}{3}$

6t dt = 2z dz

∴ t dt = $\frac{z}{3}$ dz

Here, I = $\ \int\frac{1}{\left(2x^2+5\right)\sqrt{7x^2+3}}\ dx----\left(1\right)$

$2 + 5t^2 = 2 + 5\left(\frac{z^2-7}{3}\right)$

$=\ \frac{6+5z^2-35}{3}$

$=\ \frac{5z^2-29}{3}$

$\therefore\ \ I=\ \frac{-1}{3}\int\frac{3}{\left(5z^2-29\right)\times \sqrt{z^2}}\times z\ dz$

$=\ -\int\frac{1}{5z^2-29}\ dz$

$=\ -\int\frac{1}{\left(\sqrt5z\right)^2-\ \left(\sqrt{29}\right)^2}\ dz$

$=\ -\ \frac{1}{2\sqrt{29}}\times\frac{1}{\sqrt5}\times\log{\left|\frac{\sqrt5z-\sqrt{29}}{\sqrt5z+\ \sqrt{29}}\right|}+c$

$=\ \frac{-1}{2\sqrt{145}}\log{\left|\frac{\sqrt{5\left(3t^2+7\right)}-\sqrt{29}}{\sqrt{5\left(3t^2+7\right)}+\ \sqrt{29}}\right|}+c$

$=\ \frac{-1}{2\sqrt{145}}\log{\left|\frac{\sqrt{15t^2+35}-\sqrt{29}}{\sqrt{15t^2+35}+\ \sqrt{29}}\right|}+c$

$=\ \frac{-1}{2\sqrt{145}}\log{\left|\frac{\sqrt{15t^2+35}-\ \sqrt{29}}{\sqrt{15t^2+35}+\ \sqrt{29}}\right|}+c$

$=\ \frac{-1}{2\sqrt{145}}\log{\left|\frac{\sqrt{\frac{15}{x^2}+35}-\ \sqrt{29}}{\sqrt{\frac{15}{x^2}+35}+\ \sqrt{29}}\right|}+c$

$=\ \frac{-1}{2\sqrt{145}}\log{\left|\frac{\sqrt{35x^2+15}-\ \sqrt{29x^2}}{\sqrt{35x^2+15}+\ \sqrt{29x^2}}\right|}+c$