Question

Tính các nguyên hàm sau đây :

a) \(\int\left(x+\ln x\right)x^2dx\)

b) \(\int\left(x+\sin^2x\right)\sin xdx\)

c) \(\int\left(x+e^x\right)e^{2x}dx\)

d) \(\int\left(x+\sin x\right)\dfrac{dx}{\cos^2x}\)

e) \(\int\dfrac{e^x\cos x+\left(e^x+1\right)\sin x}{e^x\sin x}dx\)

Answer 1

a) \(\int\left(x+\ln x\right)x^2\text{d}x=\int x^3\text{d}x+\int x^2\ln x\text{dx}\)

\(=\dfrac{x^4}{4}+\int x^2\ln x\text{dx}+C\) (*)

Để tính: \(\int x^2\ln x\text{dx}\) ta sử dụng công thức tính tích phân từng phần như sau:

Đặt \(\left\{{}\begin{matrix}u=\ln x\\v'=x^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u'=\dfrac{1}{x}\\v=\dfrac{1}{3}x^3\end{matrix}\right.\)

Suy ra:

\(\int x^2\ln x\text{dx}=\dfrac{1}{3}x^3\ln x-\dfrac{1}{3}\int x^2\text{dx}\)

\(=\dfrac{1}{3}x^3\ln x-\dfrac{1}{3}.\dfrac{1}{3}x^3\)

Thay vào (*) ta tính được nguyên hàm của hàm số đã cho bằng:

(*) \(=\dfrac{1}{3}x^3-\dfrac{1}{3}x^3\ln x+\dfrac{1}{9}x^3+C\)

\(=\dfrac{4}{9}x^3-\dfrac{1}{3}x^3\ln x+C\)

Answer 2

b) Đặt \(\left\{{}\begin{matrix}u=x+\sin^2x\\v'=\sin x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u'=1+2\sin x.\cos x\\v=-\cos x\end{matrix}\right.\)

Ta có:

\(\int\left(x+\sin^2x\right)\sin x\text{dx}=-\left(x+\sin^2x\right)\cos x+\int\left(1+2\sin x\cos^2x\right)\text{dx}\)

\(=-\left(x+\sin^2x\right)\cos x+\int\cos x\text{dx}+2\int\sin x.\cos^2x\text{dx}\)

\(=-\left(x+\sin^2x\right)\cos x+\sin x-2\int\cos^2x.d\left(\cos x\right)\)

\(=-\left(x+\sin^2x\right)\cos x+\sin x-2\dfrac{\cos^3x}{3}+C\)

Answer 3

c) Đặt \(\left\{{}\begin{matrix}u=x+e^x\\v'=e^{2x}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u'=1+e^x\\v=\dfrac{1}{2}e^{2x}\end{matrix}\right.\)

Ta có:

\(\int\left(x+e^x\right)e^{2x}\text{dx}=\dfrac{1}{2}\left(x+e^x\right)e^{2x}-\dfrac{1}{2}\int\left(1+e^x\right)e^{2x}\text{dx}\)

\(=\dfrac{1}{2}\left(x+e^x\right)e^{2x}-\dfrac{1}{2}\int e^{2x}\text{dx}-\dfrac{1}{2}\int e^{3x}\text{dx}\)

\(=\dfrac{1}{2}\left(x+e^x\right)e^{2x}-\dfrac{1}{2}.\dfrac{1}{2}e^{2x}-\dfrac{1}{2}.\dfrac{1}{3}e^{3x}\)

\(=\dfrac{1}{2}xe^{2x}-\dfrac{1}{4}e^{2x}+\dfrac{1}{3}e^{3x}\)

Answer 4

d) Đặt \(\left\{{}\begin{matrix}u=x+\sin x\\v'=\dfrac{1}{\cos^2x}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u'=1+\cos x\\v=\tan x\end{matrix}\right.\)

Ta có:

\(\int\left(x+\sin x\right)\dfrac{\text{dx}}{\cos^2x}=\left(x+\sin x\right)\tan x-\int\left(1+\cos x\right)\tan x\text{dx}\)

\(=x\tan x+\sin x.\tan x-\int\tan x\text{d}x-\int\cos x\tan x\text{dx}\)

\(=x\tan x+\sin x.\tan x-\int\dfrac{\sin x}{\cos x}\text{d}x-\int\cos x\dfrac{\sin x}{\cos x}\text{dx}\)

\(=x\tan x+\sin x.\tan x+\int\dfrac{d\left(\cos x\right)}{\cos x}-\int\sin x\text{dx}\)

\(=x\tan x+\sin x.\tan x+\ln\left|\cos x\right|+\cos x+C\)

Answer 5

e) \(I=\int\dfrac{e^x\cos x+\left(e^x+1\right)\sin x}{e^x\sin x}\text{dx}\)

\(=\int\left(\dfrac{\cos x}{\sin x}+1+\dfrac{1}{e^x}\right)\text{d}x\)

\(=\int\dfrac{\text{d}\left(\sin x\right)}{\sin x}+x+\int e^{-x}\text{d}x\)

\(=\ln\left|\sin x\right|+x-e^{-x}\)