Integral of $\sqrt{16x^{2}+9}$ with respect to x Find the Integral of $\sqrt{16x^{2}+9}$ with respect to x
My try: $\int\sqrt{16x^{2}+9} dx =\int(16x^{2}+9)^{1/2}dx$ then make u substitution $u=16x^{2}+9$
but then have no idea how to continue

HINT:
$$I=\int\sqrt{x^2+a^2}dx=\int dx \sqrt{x^2+a^2}-\int\left(\int dx \frac{d \sqrt{x^2+a^2}}{dx}\right)dx$$
$$=x\sqrt{x^2+a^2}-\int \frac{x^2}{\sqrt{x^2+a^2}}dx$$
$$=x\sqrt{x^2+a^2}-\int \frac{x^2+a^2-a^2}{\sqrt{x^2+a^2}}dx$$
$$=x\sqrt{x^2+a^2}-I+a^2\int \frac1{\sqrt{x^2+a^2}}dx$$
For $\displaystyle \int \frac1{\sqrt{x^2+a^2}}dx,$ use  Trigonometric substitution
Now $\displaystyle 16x^2+9=16\left(x^2+\left(\frac34\right)^2\right)$
