evaluate $int\frac{dx}{(25 + x^{2})^{2}}$.

Explanation:

Step1: Use trigonometric substitution

Let $x = 5\tan\theta$, then $dx=5\sec^{2}\theta d\theta$. And $25 + x^{2}=25+25\tan^{2}\theta = 25\sec^{2}\theta$.

Step2: Rewrite the integral

Substitute into the integral: [ \begin{align*} \int\frac{dx}{(25 + x^{2})^{2}}&=\int\frac{5\sec^{2}\theta d\theta}{(25\sec^{2}\theta)^{2}}\ &=\int\frac{5\sec^{2}\theta d\theta}{625\sec^{4}\theta}\ &=\frac{1}{125}\int\cos^{2}\theta d\theta \end{align*} ]

Step3: Use the double - angle formula

Since $\cos^{2}\theta=\frac{1 + \cos(2\theta)}{2}$, the integral becomes $\frac{1}{125}\int\frac{1+\cos(2\theta)}{2}d\theta=\frac{1}{250}\int(1+\cos(2\theta))d\theta$.

Step4: Integrate term - by - term

$\frac{1}{250}\int(1+\cos(2\theta))d\theta=\frac{1}{250}\left(\theta+\frac{1}{2}\sin(2\theta)\right)+C$.

Step5: Back - substitute

Since $x = 5\tan\theta$, then $\theta=\arctan\left(\frac{x}{5}\right)$ and $\sin(2\theta)=\frac{2\tan\theta}{1 + \tan^{2}\theta}=\frac{2\frac{x}{5}}{1+\frac{x^{2}}{25}}=\frac{10x}{25 + x^{2}}$. [ \begin{align*} \frac{1}{250}\left(\theta+\frac{1}{2}\sin(2\theta)\right)+C&=\frac{1}{250}\left(\arctan\left(\frac{x}{5}\right)+\frac{1}{2}\cdot\frac{10x}{25 + x^{2}}\right)+C\ &=\frac{1}{250}\arctan\left(\frac{x}{5}\right)+\frac{x}{50(25 + x^{2})}+C \end{align*} ]

Answer:

$\frac{1}{250}\arctan\left(\frac{x}{5}\right)+\frac{x}{50(25 + x^{2})}+C$