Nguyên hàm 4

Tìm nguyên hàm:  $I=\displaystyle\int x^2\sqrt{x^2+9}dx$

Đặt $u=x\implies u'=1$; $v'=x\sqrt{x^2+9}$ chọn $ v=\frac{1}{3}(x^2+9)\sqrt{x^2+9}$ $$I=\frac{1}{3}x(x^2+9)\sqrt{x^2+9}-\frac{1}{3}\int(x^2+9)\sqrt{x^2+9}dx$$ $$ \iff \frac{4}{3}I=\frac{1}{3}x(x^2+9)\sqrt{x^2+9}-3\int \sqrt{x^2+9}dx$$ Mà $\displaystyle\int \sqrt{x^2+9}dx=x\sqrt{x^2+9}-\displaystyle\int \frac{x^2}{\sqrt{x^2+9}}dx$

 $\displaystyle\int \sqrt{x^2+9}dx=x\sqrt{x^2+9}-\displaystyle\int \sqrt{x^2+9}dx+9\displaystyle\int \frac{dx}{\sqrt{x^2+9}}$

 Vì $\left(\ln\left(x+\sqrt{x^2+9}\right)\right)'=\dfrac{1}{\sqrt{x^2+9}}$ $$2\displaystyle\int \sqrt{x^2+9}dx=x\sqrt{x^2+9}+9\ln\left(x+\sqrt{x^2+9}\right)$$ Vậy $I=\dfrac{1}{4}x(x^2+9)\sqrt{x^2+9}-\dfrac{9}{8}x\sqrt{x^2+9}-\dfrac{81}{8}\ln\left(x+\sqrt{x^2+9}\right)+C$