Đáp án: $\displaystyle\int \sqrt{x^8+x^6}dx=\dfrac{1}{15}\left(3x^2-2\right)\left(x^2+1\right)^{\dfrac{3}{2}}+C$
Giải thích các bước giải:
$I=\displaystyle\int \sqrt{x^8+x^6}dx$
$=\displaystyle\int x^3\sqrt{x^2+1}dx$
Đặt $x^2+1=t\to 2xdx=dt$
$\to I=\displaystyle\int \dfrac12x^2\sqrt{x^2+1}\cdot 2xdx$
$\to I=\displaystyle\int \dfrac12\left(t-1\right)\sqrt{t}dt$
$\to I=\dfrac12\displaystyle\int t\sqrt{t}-\sqrt{t}dt$
$\to I=\dfrac12\displaystyle\int t^{\dfrac32}-t^{\dfrac12}dt$
$\to I=\dfrac12\left(\dfrac25t^{\dfrac52}-\dfrac23t^{\dfrac32}\right)$
$\to I=\dfrac{1}{2}\left(\dfrac{2}{5}\left(x^2+1\right)^{\dfrac{5}{2}}-\dfrac{2}{3}\left(x^2+1\right)^{\dfrac{3}{2}}\right)+C$
$\to I=\dfrac{1}{15}\left(3x^2-2\right)\left(x^2+1\right)^{\dfrac{3}{2}}+C$
