Question

Chapter 5, Review Exercises, Question 013 x Evaluate the integral ✓ dx by making the substitution u = x - 6. (1x2 - 6x4 - 12x2 Enter the exact answer. Ir-ore- dx = ? Edit +C (x2 - 6V4 – 12x2

Answer 1

du (2-6) x4 – 12x2

$\small =\int \frac{x}{(x^2-6)\sqrt{(x^2)^2-(2)(6)x^2+(6)^2-(6)^2}}dx$

$\small =\int \frac{x}{(x^2-6)\sqrt{(x^2-6)^2-36}}dx$$\small (\because (a-b)^2=a^2-2ab+b^2)$

Substitute $\small u=x^2-6\rightarrow du=(2x-0)dx$

$\small \Rightarrow xdx=\frac{du}{2}$

So, the integration becomes

$\small \int \frac{1}{u\sqrt{u^2-36}}(\frac{du}{2})$

$\small =\frac{1}{2}\int \frac{1}{u\sqrt{u^2-36}}du$

Substitute $\small v=\sqrt{u^2-36} \Rightarrow \frac{\mathrm{d} v}{\mathrm{d} u}=\frac{\mathrm{d} }{\mathrm{d} u}(\sqrt{u^2-36})$

$\small \Rightarrow \frac{\mathrm{d} v}{\mathrm{d} u}=\frac{\mathrm{d} }{\mathrm{d} (u^2-36)}(\sqrt{u^2-36})\times \frac{\mathrm{d} }{\mathrm{d} u}(u^2-36)$ (Chain Rule is applied here.)

$\small \Rightarrow \frac{\mathrm{d} v}{\mathrm{d} u}=\frac{1}{2}(u^2-36)^{1/2-1}\times(2u)$$\small ((x^n)'=nx^{n-1})$

$\small \Rightarrow \frac{\mathrm{d} v}{\mathrm{d} u}=\frac{u}{\sqrt{u^2-36}}$

du VU? - 36 -du = u

So, integration becomes

$\small \frac{1}{2}\int \frac{1}{u\sqrt{u^2-36}}(\frac{\sqrt{u^2-36}}{u}dv)$

$\small =\frac{1}{2}\int \frac{1}{u^2}dv$

$\small =\frac{1}{2}\int \frac{1}{v^2+36}dv$$\small (\because v^2=u^2-36;From \: Substitution)$

$\small =\frac{1}{2}\int \frac{1}{v^2+(6)^2}dv$

$\small =\frac{1}{2}(\frac{1}{6}tan^{-1}(\frac{v}{6}))+c$$\small (\int \frac{1}{x^2+a^2}=\frac{1}{a}tan^{-1}(\frac{x}{a});a\rightarrow constant)$

$\small =\frac{1}{12}tan^{-1}(\frac{v}{6})+c$

$\small =\frac{1}{12}tan^{-1}(\frac{\sqrt{u^2-36}}{6})+c$

We have substituted
=N 7
$\small u=x^2-6\rightarrow u^2=(x^2-6)^2$

$\small \Rightarrow u^2=(x^4-12x^2+36)$

$\small \Rightarrow u^2-36=(x^4-12x^2)$

$\small \Rightarrow \sqrt{u^2-36}=\sqrt{x^4-12x^2}$

$\small \therefore \int \frac{x}{(x^2-6)\sqrt{x^4-12x^2}}dx=\mathbf{\frac{1}{12}tan^{-1}(\frac{\sqrt{x^4-12x^2}}{6})}+C$