họ nguyên hàm của hàm số x^3(x^2+1)^2019 là?

$\displaystyle\int x^3(x^2+1)^{2019} dx\\ =\displaystyle\int ((x^3+x)(x^2+1)^{2019}- x(x^2+1)^{2019})dx\\ =\displaystyle\int (x(x^2+1)^{2020}- x(x^2+1)^{2019})dx\\ =\dfrac{1}{2}\displaystyle\int ((x^2+1)^{2020}- (x^2+1)^{2019})d(x^2+1)\\ =\dfrac{1}{4042}.(x^2+1)^{2021}-\dfrac{1}{4040}.(x^2+1)^{2020}+C$

$C2:$ Đặt $t=x^2+1$

$=>x=\sqrt{t-1}\\ dx=\dfrac{1}{2\sqrt{t-1}}dt\\ I=\displaystyle\int \sqrt{t-1}(t-1)t^{2019} \dfrac{1}{2\sqrt{t-1}}dt\\ =\dfrac{1}{2}\displaystyle\int(t-1)t^{2019} dt\\ =\dfrac{1}{2}\displaystyle\int(t^{2020}-t^{2019}) dt\\ =\dfrac{1}{4042}.t^{2021}-\dfrac{1}{4040}.t^{2020}+C\\ =\dfrac{1}{4042}.(x^2+1)^{2021}-\dfrac{1}{4040}.(x^2+1)^{2020}+C$

Đáp án:

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