Question

use any method to evaluate the integral.
int\frac{xmathrm{d}x}{sqrt{100x^{2}-1}}
int\frac{xmathrm{d}x}{sqrt{100x^{2}-1}}=square

Explanation:

Step1: Use substitution

Let $u = 100x^{2}-1$, then $du=200x\ dx$, and $x\ dx=\frac{1}{200}du$.

Step2: Rewrite the integral

The integral $\int\frac{x\ dx}{\sqrt{100x^{2}-1}}$ becomes $\int\frac{\frac{1}{200}du}{\sqrt{u}}=\frac{1}{200}\int u^{-\frac{1}{2}}du$.

Step3: Integrate using power - rule

The power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$). For $n=-\frac{1}{2}$, we have $\frac{1}{200}\int u^{-\frac{1}{2}}du=\frac{1}{200}\times\frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C$.

Step4: Simplify and substitute back

$\frac{1}{200}\times\frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C=\frac{1}{100}\sqrt{u}+C$. Substituting $u = 100x^{2}-1$ back, we get $\frac{1}{100}\sqrt{100x^{2}-1}+C$.

Answer:

$\frac{1}{100}\sqrt{100x^{2}-1}+C$