Question

$\int \frac{dx}{(1+x^2)\sqrt{p^2+q^2(ta{n}^{-1}x{)}^2}}=$

• A. $\frac{1}{q}log[qta{n}^{-1}x+\sqrt{p^2+q^2(ta{n}^{-1}x{)}^2}]+c$
• B. $log[qta{n}^{-1}x+\sqrt{p^2+q^2(ta{n}^{-1}x{)}^2}]+c$
• C. $\frac{2}{3q}(p^2+q^{2}ta{n}^{-1}x{)}^{3/2}+c$
• D. None of these

Answer 1

The correct option is A $\frac{1}{q}log[qta{n}^{-1}x+\sqrt{p^2+q^2(ta{n}^{-1}x{)}^2}]+c$

Putting $qta{n}^{-1}x=t\Rightarrow \frac{q}{1+x^2}dx=dt\Rightarrow \frac{1}{1+x^2}dx=\frac{dt}{q}$
$\Rightarrow \int \frac{dx}{(1+x^2)\sqrt{p^2+q^2(ta{n}^{-1}x{)}^2}}=\frac{1}{q}\int \frac{dt}{\sqrt{p^2+t^2}}$
$=\frac{1}{q}log[qta{n}^{-1}x+\sqrt{p^2+q^2(ta{n}^{-1}x{)}^2}]+c$