Tìm nguyên hàm $\int {\frac{(x+1)^{2010}}{(3x+1)^{2012}}} \, dx$ 04/09/2021 Bởi Adalynn Tìm nguyên hàm $\int {\frac{(x+1)^{2010}}{(3x+1)^{2012}}} \, dx$
Ta có $\int \dfrac{(x+1)^{2010}}{(3x+1)^{2012}} dx= \int \left( \dfrac{x+1}{3x+1} \right)^{2010} . \dfrac{1}{(3x+1)^2} dx$ Đặt $t = \dfrac{x+1}{3x+1}$. Khi đó $dt = d\left(\dfrac{x+1}{3x+1}\right)$ $= \left(\dfrac{x+1}{3x+1}\right)’ dx$ $= \dfrac{-2}{(3x+1)^2}dx$ Vậy $\dfrac{1}{(3x+1)^2} = -\dfrac{dt}{2}$. Thay vào ta có $\int \dfrac{(x+1)^{2010}}{(3x+1)^{2012}} dx = -\dfrac{1}{2}\int t^{2010} . dt$ $= -\dfrac{1}{2} \dfrac{t^{2011}}{2011} + c$ $= -\dfrac{1}{4022} \dfrac{(x+1)^{2011}}{(3x+1)^{2011}} + c$. Do đó $\int \dfrac{(x+1)^{2010}}{(3x+1)^{2012}} dx = -\dfrac{(x+1)^{2011}}{4022(3x+1)^{2011}} + c$. Bình luận
Ta có
$\int \dfrac{(x+1)^{2010}}{(3x+1)^{2012}} dx= \int \left( \dfrac{x+1}{3x+1} \right)^{2010} . \dfrac{1}{(3x+1)^2} dx$
Đặt $t = \dfrac{x+1}{3x+1}$. Khi đó
$dt = d\left(\dfrac{x+1}{3x+1}\right)$
$= \left(\dfrac{x+1}{3x+1}\right)’ dx$
$= \dfrac{-2}{(3x+1)^2}dx$
Vậy
$\dfrac{1}{(3x+1)^2} = -\dfrac{dt}{2}$.
Thay vào ta có
$\int \dfrac{(x+1)^{2010}}{(3x+1)^{2012}} dx = -\dfrac{1}{2}\int t^{2010} . dt$
$= -\dfrac{1}{2} \dfrac{t^{2011}}{2011} + c$
$= -\dfrac{1}{4022} \dfrac{(x+1)^{2011}}{(3x+1)^{2011}} + c$.
Do đó
$\int \dfrac{(x+1)^{2010}}{(3x+1)^{2012}} dx = -\dfrac{(x+1)^{2011}}{4022(3x+1)^{2011}} + c$.