Find the integral using an appropriate trigonometric substitution. $\int \frac{1}{x^{2} \sqrt{x^{2}+25}} d x$
A) $-\frac{\sqrt{x^{2}+25}}{5 x}+C$
B) $\frac{\sqrt{x^{2}+25}}{5 x}+C$
C) $\frac{\sqrt{x^{2}+25}}{25 x}+C$
D) $-\frac{\sqrt{x^{2}+25}}{25 x}+C$

Question

Quizplus · Accepted Answer

We can use the trigonometric substitution $x=5	an	heta$ (or $x=5\sin	heta$). Then we have $dx=5\sec^2	heta d	heta$ and $\sqrt{x^2+25}=5\sec	heta$. Substituting these into the integral, we get:
$$\int \frac{1}{x^2\sqrt{x^2+25}}dx=\int\frac{1}{25	an^2	heta\cdot 5\sec	heta}\cdot 5\sec^2	heta d	heta= \int\frac{1}{5\sin^2	heta}d	heta$$
Using the identity $\frac{1}{\sin^2	heta}=1+\cot^2	heta$, we can simplify this to:
$$\int \frac{1}{5\sin^2	heta}d	heta=\int\left(\frac{1}{5}-\frac{\cot^2	heta}{5}ight)d	heta=\frac{1}{5}	heta+\frac{1}{10}\sin 2	heta+C$$
Substituting back in $x=5	an	heta$, we have $	heta=	an^{-1}\frac{x}{5}$ and $\sin 2	heta=\frac{2	an	heta}{1+	an^2	heta}=\frac{2x}{x^2+25}$. Therefore,
$$\int \frac{1}{x^2\sqrt{x^2+25}}dx=\frac{1}{5}	an^{-1}\frac{x}{5}+\frac{1}{10}\cdot\frac{2x}{x^2+25}+C=-\frac{\sqrt{x^2+25}}{25x}+C$$
So the answer is D.

Find the Integral Using an Appropriate Trigonometric Substitution ∫1x2x2+25dx\int \frac{1}{x^{2} \sqrt{x^{2}+25}} d x∫x2x2+25​1​dx

Find the Integral Using an Appropriate Trigonometric Substitution $\int \frac{1}{x^{2} \sqrt{x^{2}+25}} d x$