Tìm nguyên hàm : I = $\int\limits^ {}\,\sqrt[]{x^2 + 9} dx$

Đáp án:

Giải thích các bước giải:

$\int \sqrt{x^2+9}dx$

$=\int \:9\sec ^3\left(u\right)du

$=9\left(\dfrac{\sec ^2\left(u\right)\sin \left(u\right)}{2}+\dfrac{1}{2}\cdot \int \sec \left(u\right)du\right)$

$=9\left(\dfrac{\sec ^2\left(u\right)\sin \left(u\right)}{2}+\dfrac{1}{2}\ln \left|\tan \left(u\right)+\sec \left(u\right)\right|\right)$

$=9\left(\dfrac{\sec ^2\left(\arctan \left(\dfrac{1}{3}x\right)\right)\sin \left(\arctan \left(\dfrac{1}{3}x\right)\right)}{2}+\frac{1}{2}\ln \left|\tan \left(\arctan \left(\dfrac{1}{3}x\right)\right)+\sec \left(\arctan \left(\dfrac{1}{3}x\right)\right)\right|\right)$

$=9\left(\dfrac{1}{18}x\sqrt{9+x^2}+\dfrac{1}{2}\ln \left(\dfrac{\left|x+\sqrt{9+x^2}\right|}{3}\right)\right)+C$