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(B) To evaluate the integral $I = \int \sqrt{x^2+3x} \, dx$,we complete the square inside the square root:$x^2 + 3x = (x + \frac{3}{2})^2 - (\frac{3}{

Vedclass · Accepted Answer

$\frac{2x+3}{4} \sqrt{x^2+3x} - \frac{9}{8} \log \left| x + \frac{3}{2} + \sqrt{x^2+3x} \right| + c$

Vedclass · Answer

(B) To evaluate the integral $I = \int \sqrt{x^2+3x} \, dx$,we complete the square inside the square root:$x^2 + 3x = (x + \frac{3}{2})^2 - (\frac{3}{2})^2 = (x + \frac{3}{2})^2 - \frac{9}{4}$.Now,the integral becomes $I = \int \sqrt{(x + \frac{3}{2})^2 - (\frac{3}{2})^2} \, dx$.Using the standard formula $\int \sqrt{t^2 - a^2} \, dt = \frac{t}{2} \sqrt{t^2 - a^2} - \frac{a^2}{2} \log |t + \sqrt{t^2 - a^2}| + c$,where $t = x + \frac{3}{2}$ and $a = \frac{3}{2}$:$I = \frac{x + \frac{3}{2}}{2} \sqrt{(x + \frac{3}{2})^2 - \frac{9}{4}} - \frac{9/4}{2} \log |(x + \frac{3}{2}) + \sqrt{(x + \frac{3}{2})^2 - \frac{9}{4}}| + c$.Simplifying this,we get:$I = \frac{2x+3}{4} \sqrt{x^2+3x} - \frac{9}{8} \log |x + \frac{3}{2} + \sqrt{x^2+3x}| + c$.

$\int \sqrt{x^2+3x} \, dx =$

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