evaluate the integral using a suitable trigonometric substitution.\n int \frac{sqrt{49 + 49x^{2}}}{8x} dx

Explanation:

Step1: Simplify the integrand

First, factor out 49 from the square - root: $\sqrt{49 + 49x^{2}}=\sqrt{49(1 + x^{2})}=7\sqrt{1 + x^{2}}$. So the integral becomes $\frac{7}{8}\int\frac{\sqrt{1 + x^{2}}}{x}dx$.

Step2: Make a trigonometric substitution

Let $x = \tan\theta$, then $dx=\sec^{2}\theta d\theta$. And $\sqrt{1 + x^{2}}=\sqrt{1+\tan^{2}\theta}=\sec\theta$. The integral $\frac{7}{8}\int\frac{\sqrt{1 + x^{2}}}{x}dx$ becomes $\frac{7}{8}\int\frac{\sec\theta}{\tan\theta}\sec^{2}\theta d\theta=\frac{7}{8}\int\frac{\sec^{3}\theta}{\tan\theta}d\theta$. Since $\sec\theta=\frac{1}{\cos\theta}$ and $\tan\theta=\frac{\sin\theta}{\cos\theta}$, we have $\frac{7}{8}\int\frac{\frac{1}{\cos^{3}\theta}}{\frac{\sin\theta}{\cos\theta}}d\theta=\frac{7}{8}\int\frac{1}{\sin\theta\cos^{2}\theta}d\theta$.

Step3: Rewrite the integrand in terms of sine and cosine

$\frac{1}{\sin\theta\cos^{2}\theta}=\frac{\sin\theta}{\sin^{2}\theta\cos^{2}\theta}=\frac{\sin\theta}{(1 - \cos^{2}\theta)\cos^{2}\theta}$. Let $u = \cos\theta$, then $du=-\sin\theta d\theta$. The integral becomes $-\frac{7}{8}\int\frac{1}{(1 - u^{2})u^{2}}du$.

Step4: Decompose the fraction into partial - fractions

We decompose $\frac{1}{(1 - u^{2})u^{2}}=\frac{1}{(1 - u)(1 + u)u^{2}}=\frac{A}{u}+\frac{B}{u^{2}}+\frac{C}{1 - u}+\frac{D}{1 + u}$. After finding the coefficients $A = 0$, $B = 1$, $C=\frac{1}{2}$, $D=\frac{1}{2}$, the integral $-\frac{7}{8}\int\frac{1}{(1 - u^{2})u^{2}}du=-\frac{7}{8}\int\left(\frac{1}{u^{2}}+\frac{1}{2(1 - u)}+\frac{1}{2(1 + u)}\right)du$.

Step5: Integrate term - by - term

$-\frac{7}{8}\int\left(\frac{1}{u^{2}}+\frac{1}{2(1 - u)}+\frac{1}{2(1 + u)}\right)du=-\frac{7}{8}\left(-\frac{1}{u}-\frac{1}{2}\ln|1 - u|+\frac{1}{2}\ln|1 + u|\right)+C$.

Step6: Back - substitute

Since $u = \cos\theta$ and $x=\tan\theta$, $\cos\theta=\frac{1}{\sqrt{1 + x^{2}}}$. The integral $\frac{7}{8}\left(\sqrt{1 + x^{2}}+\frac{1}{2}\ln\left|\frac{\sqrt{1 + x^{2}}-1}{\sqrt{1 + x^{2}}+1}\right|\right)+C$.

Answer:

$\frac{7}{8}\left(\sqrt{1 + x^{2}}+\frac{1}{2}\ln\left|\frac{\sqrt{1 + x^{2}}-1}{\sqrt{1 + x^{2}}+1}\right|\right)+C$