Question

tìm nguyên hàm của hàm số sau bằng phương pháp nguyên hàm từng phần:

\(\int x\left(e^{2x}+\sqrt[3]{x+1}\right)dx\)

tìm nguyên hàm:

\(\int\frac{2x+3}{x-2}dx\)

Answer 1

\(I=\int\left(xe^{2x}+x\sqrt[3]{x+1}\right)dx=\int xe^{2x}dx+\int x\sqrt[3]{x+1}dx=I_1+I_2\)

Xét \(I_1=\int xe^{2x}dx\Rightarrow\left\{{}\begin{matrix}u=x\\dv=e^{2x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\frac{1}{2}e^{2x}\end{matrix}\right.\)

\(\Rightarrow I_1=\frac{1}{2}xe^{2x}-\int\frac{1}{2}e^{2x}dx=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}+C_1\)

Xét \(I_2=\int x\sqrt[3]{x+1}dx\)

Đặt \(\sqrt[3]{x+1}=t\Rightarrow x=t^3-1\Rightarrow dx=3t^2dt\)

\(\Rightarrow I_2=\int\left(t^3-1\right).t.3t^2dt=\int\left(3t^6-3t^3\right)dt=\frac{3}{7}t^7-\frac{3}{4}t^4+C_2\)

\(=\frac{3}{7}\sqrt[3]{\left(x+1\right)^7}-\frac{3}{4}\sqrt[3]{\left(x+1\right)^4}+C_2\)

\(\Rightarrow I=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}+\frac{3}{7}\sqrt[3]{\left(x+1\right)^7}-\frac{3}{4}\sqrt[3]{\left(x+1\right)^4}+C\)

Answer 2

\(I=\int\frac{2x+3}{x-2}dx\)

\(=\int\frac{2\left(x-2\right)+7}{x-2}dx\)

\(=\int\left(2+\frac{7}{x-2}\right)dx=2x+7ln\left|x-2\right|+C\)