Câu 2)

Đặt \(\left\{\begin{matrix} u=\ln ^2x\\ dv=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2\frac{\ln x}{x}dx\\ v=\frac{x^3}{3}\end{matrix}\right.\Rightarrow I=\frac{x^3}{3}\ln ^2x-\frac{2}{3}\int x^2\ln xdx\)

Đặt \(\left\{\begin{matrix} k=\ln x\\ dt=x^2dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} dk=\frac{dx}{x}\\ t=\frac{x^3}{3}\end{matrix}\right.\Rightarrow \int x^2\ln xdx=\frac{x^3\ln x}{3}-\int \frac{x^2}{3}dx=\frac{x^3\ln x}{3}-\frac{x^3}{9}+c\)

Do đó \(I=\frac{x^3\ln^2x}{3}-\frac{2}{9}x^3\ln x+\frac{2}{27}x^3+c\)

Câu 3:

\(I=\int\frac{2}{\cos 2x-7}dx=-\int\frac{2}{2\sin^2x+6}dx=-\int\frac{dx}{\sin^2x+3}\)

Đặt \(t=\tan\frac{x}{2}\Rightarrow \left\{\begin{matrix} \sin x=\frac{2t}{t^2+1}\\ dx=\frac{2dt}{t^2+1}\end{matrix}\right.\)

\(\Rightarrow I=-\int \frac{2dt}{(t^2+1)\left ( \frac{4t^2}{(t^2+1)^2}+3 \right )}=-\int\frac{2(t^2+1)dt}{3t^4+10t^2+3}=-\int \frac{2d\left ( t-\frac{1}{t} \right )}{3\left ( t-\frac{1}{t} \right )^2+16}=\int\frac{2dk}{3k^2+16}\)

Đặt \(k=\frac{4}{\sqrt{3}}\tan v\). Đến đây dễ dàng suy ra \(I=\frac{-1}{2\sqrt{3}}v+c\)

Câu 6)

\(I=-\int \frac{\left ( 1-\frac{1}{x^2} \right )dx}{x^2+2+\frac{1}{x^2}}=-\int \frac{d\left ( x+\frac{1}{x} \right )}{\left ( x+\frac{1}{x} \right )^2}=-\frac{1}{x+\frac{1}{x}}+c=-\frac{x}{x^2+1}+c\)

Câu 8)

\(I=\int \ln \left(\frac{x+1}{x-1}\right)dx=\int \ln (x+1)dx-\int \ln (x-1)dx\)

\(\Leftrightarrow I=\int \ln (x+1)d(x+1)-\int \ln (x-1)d(x-1)\)

Xét \(\int \ln tdt\) ta có:

Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln tdt=t\ln t-\int dt=t\ln t-t+c\)

\(\Rightarrow I=(x+1)\ln (x+1)-(x+1)-(x-1)\ln (x-1)+x-1+c\)

\(\Leftrightarrow I=(x+1)\ln(x+1)-(x-1)\ln(x-1)+c\)

\(\int\dfrac{lnx}{x\left(2ln^2x-1\right)^3}dx\)

\(t=2ln^2x-1\Rightarrow dt=\dfrac{4}{x}lnxdx\Rightarrow dx=\dfrac{x.dt}{4lnx}\)

\(\Rightarrow\int\dfrac{lnx}{x\left(2ln^2x-1\right)^3}dx=\int\dfrac{lnx}{x\left(2ln^2x-1\right)^3}.\dfrac{xdt}{4lnx}=\dfrac{1}{4}\int\dfrac{dt}{t^3}=\dfrac{1}{4}.\left(-\dfrac{1}{2}\right).t^{-2}=-\dfrac{1}{8\sqrt{2ln^2x-1}}\)

\(\int e^x\left(2-x\right)dx\)

\(\left\{{}\begin{matrix}u=2-x\\dv=e^xdx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=-dx\\v=e^x\end{matrix}\right.\)

\(\Rightarrow\int e^x\left(2-x\right)dx=e^x\left(2-x\right)+\int e^xdx=e^x\left(2-x\right)+e^x\)

\(\int\limits^3_{-1}f\left(\left|x\right|\right)dx=\int\limits^0_{-1}f\left(\left|x\right|\right)dx+\int\limits^1_0f\left(\left|x\right|\right)dx+\int\limits^3_1f\left(\left|x\right|\right)dx\)

\(=\int\limits^0_{-1}f\left(-x\right)dx+\int\limits^1_0f\left(x\right)dx+\int\limits^3_1f\left(x\right)dx\)

\(=\int\limits^1_0f\left(x\right)dx+\int\limits^1_0f\left(x\right)dx+\int\limits^3_1f\left(x\right)dx\)

\(=3+3+6=12\)

a) Áp dụng phương pháp tìm nguyên hàm từng phần:

Đặt u= ln(1+x)

dv= xdx

=> ,

Ta có: ∫xln(1+x)dx =

=

b) Cách 1: Tìm nguyên hàm từng phần hai lần:

Đặt u= (x²+2x -1) và dv=e^xdx

Suy ra du = (2x+2)dx, v = e^x

. Khi đó:

∫(x²+2x - 1)e^xdx = (x²+2x - 1)e^xdx - ∫(2x+2)e^xdx

Đặt : u=2x+2; dv=e^xdx

=> du = 2dx ;v=e^x

Khi đó:∫(2x+2)e^xdx = (2x+2)e^x - 2∫e^xdx = e^x(2x+2) – 2e^x+C

Vậy

∫(x²+2x+1)e^xdx = e^x(x²-1) + C

Cách 2: HD: Ta tìm ∫(x²-1)e^xdx. Đặt u = x²-1 và dv=e^xdx.

Đáp số : e^x(x²-1) + C

c) Đáp số:

HD: Đặt u=x ; dv = sin(2x+1)dx

d) Đáp số : (1-x)sinx - cosx +C.

HD: Đặt u = 1 - x ;dv = cosxdx