a. \(\int\dfrac{x^3}{x-2}dx=\int\left(x^2+2x+4+\dfrac{8}{x-2}\right)dx=\dfrac{1}{3}x^3+x^2+4x+8ln\left|x-2\right|+C\)

b. \(\int\dfrac{dx}{x\sqrt{x^2+1}}=\int\dfrac{xdx}{x^2\sqrt{x^2+1}}\)

Đặt \(\sqrt{x^2+1}=u\Rightarrow x^2=u^2-1\Rightarrow xdx=udu\)

\(I=\int\dfrac{udu}{\left(u^2-1\right)u}=\int\dfrac{du}{u^2-1}=\dfrac{1}{2}\int\left(\dfrac{1}{u-1}-\dfrac{1}{u+1}\right)du=\dfrac{1}{2}ln\left|\dfrac{u-1}{u+1}\right|+C\)

\(=\dfrac{1}{2}ln\left|\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+1}+1}\right|+C\)

c. \(\int\left(\dfrac{5}{x}+\sqrt{x^3}\right)dx=\int\left(\dfrac{5}{x}+x^{\dfrac{3}{2}}\right)dx=5ln\left|x\right|+\dfrac{2}{5}\sqrt{x^5}+C\)

d. \(\int\dfrac{x\sqrt{x}+\sqrt{x}}{x^2}dx=\int\left(x^{-\dfrac{1}{2}}+x^{-\dfrac{3}{2}}\right)dx=2\sqrt{x}-\dfrac{1}{2\sqrt{x}}+C\)

e. \(\int\dfrac{dx}{\sqrt{1-x^2}}=arcsin\left(x\right)+C\)

Cậu sống ở đâu hở ? Lấy đâu ra toán khó thế ?

Câu 3 sửa \(\int\limits_1^{3/2} \)

Khó như thế này thì ai mà làm nổi

a)

Đặt \(u=\sqrt{x-3}\Rightarrow x=u^2+3\)

\(I_1=\int (2x-3)\sqrt{x-3}dx=\int (2u^2+3)ud(u^2+3)=2\int (2u^2+3)u^2du\)

\(\Leftrightarrow I_1=4\int u^4du+6\int u^2du=\frac{4u^5}{5}+2u^3+c\)

b)

\(I_2=\int \frac{xdx}{\sqrt{(x^2+1)^3}}=\frac{1}{2}\int \frac{d(x^2+1)}{\sqrt{(x^2+1)^2}}\)

Đặt \(u=\sqrt{x^2+1}\). Khi đó:

\(I_2=\frac{1}{2}\int \frac{d(u^2)}{u^3}=\int \frac{udu}{u^3}=\int \frac{du}{u^2}=\frac{-1}{u}+c\)

c)

\(I_3=\int \frac{e^xdx}{e^x+e^{-x}}=\int \frac{e^{2x}dx}{e^{2x}+1}=\frac{1}{2}\int\frac{d(e^{2x}+1)}{e^{2x}+1}\)

\(\Leftrightarrow I_3=\frac{1}{3}\ln |e^{2x}+1|+c=\frac{1}{2}\ln|u|+c\)

d)

\(I_4=\int \frac{dx}{\sin x-\sin a}=\int \frac{dx}{2\cos \left ( \frac{x+a}{2} \right )\sin \left ( \frac{x-a}{2} \right )}\)

\(\Leftrightarrow I_4=\frac{1}{\cos a}\int \frac{\cos \left ( \frac{x+a}{2}-\frac{x-a}{2} \right )dx}{2\cos \left ( \frac{x+a}{2} \right )\sin \left ( \frac{x-a}{2} \right )}=\frac{1}{\cos a}\int \frac{\cos \left ( \frac{x-a}{2} \right )dx}{2\sin \left ( \frac{x-a}{2} \right )}+\frac{1}{\cos a}\int \frac{\sin \left ( \frac{x+a}{2} \right )dx}{2\cos \left ( \frac{x+a}{2} \right )}\)

\(\Leftrightarrow I_4=\frac{1}{\cos a}\left ( \ln |\sin \frac{x-a}{2}|-\ln |\cos \frac{x+a}{2}| \right )+c\)

e)

Đặt \(t=\sqrt{x}\Rightarrow x=t^2\)

\(I_5=\int t\sin td(t^2)=2\int t^2\sin tdt\)

Đặt \(\left\{\begin{matrix} u=t^2\\ dv=\sin tdt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2tdt\\ v=-\cos t\end{matrix}\right.\)

\(\Rightarrow I_5=-2t^2\cos t+4\int t\cos tdt\)

Tiếp tục nguyên hàm từng phần \(\Rightarrow \int t\cos tdt=t\sin t+\cos t+c\)

\(\Rightarrow I_5=-2t^2\cos t+4t\sin t+4\cos t+c\)

g)

Có \(I_6=\int x\ln \left ( \frac{x}{x+1} \right )dx=\int x\ln xdx-\int x\ln (x+1)dx\)

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

\(\Leftrightarrow \int x\ln xdx=\frac{x^2\ln x}{2}-\frac{x^2}{4}+c\)

Tương tự, \(\int x\ln (x+1)dx=\frac{x^2\ln (x+1)}{2}-\int \frac{x^2}{2(x+1)}dx\)

\(=\frac{x^2\ln (x+1)}{2}-\frac{x^2}{4}+\frac{x}{2}-\frac{\ln (x+1)}{2}+c\)

Suy ra \(I_5=\frac{x^2}{2}\ln \frac{x}{x+1}+\frac{1}{2}\ln|x+1|-\frac{x}{2}+c\)

Không phải tất cả các câu đều dùng nguyên hàm từng phần được đâu nhé, 1 số câu phải dùng đổi biến, đặc biệt những câu liên quan đến căn thức thì đừng dại mà nguyên hàm từng phần (vì càng nguyên hàm từng phần biểu thức nó càng phình to ra chứ không thu gọn lại, vĩnh viễn không ra kết quả đâu)

a/ \(I=\int\frac{9x^2}{\sqrt{1-x^3}}dx\)

Đặt \(u=\sqrt{1-x^3}\Rightarrow u^2=1-x^3\Rightarrow2u.du=-3x^2dx\)

\(\Rightarrow9x^2dx=-6udu\)

\(\Rightarrow I=\int\frac{-6u.du}{u}=-6\int du=-6u+C=-6\sqrt{1-x^3}+C\)

b/ Đặt \(u=1+\sqrt{x}\Rightarrow du=\frac{dx}{2\sqrt{x}}\Rightarrow2du=\frac{dx}{\sqrt{x}}\)

\(\Rightarrow I=\int\frac{2du}{u^3}=2\int u^{-3}du=-u^{-2}+C=-\frac{1}{u^2}+C=-\frac{1}{\left(1+\sqrt{x}\right)^2}+C\)

c/ Đặt \(u=\sqrt{2x+3}\Rightarrow u^2=2x\Rightarrow\left\{{}\begin{matrix}x=\frac{u^2}{2}\\dx=u.du\end{matrix}\right.\)

\(\Rightarrow I=\int\frac{u^2.u.du}{2u}=\frac{1}{2}\int u^2du=\frac{1}{6}u^3+C=\frac{1}{6}\sqrt{\left(2x+3\right)^3}+C\)

d/ Đặt \(u=\sqrt{1+e^x}\Rightarrow u^2-1=e^x\Rightarrow2u.du=e^xdx\)

\(\Rightarrow I=\int\frac{\left(u^2-1\right).2u.du}{u}=2\int\left(u^2-1\right)du=\frac{2}{3}u^3-2u+C\)

\(=\frac{2}{3}\sqrt{\left(1+e^x\right)^2}-2\sqrt{1+e^x}+C\)

e/ Đặt \(u=\sqrt[3]{1+lnx}\Rightarrow u^3=1+lnx\Rightarrow3u^2du=\frac{dx}{x}\)

\(\Rightarrow I=\int u.3u^2du=3\int u^3du=\frac{3}{4}u^4+C=\frac{3}{4}\sqrt[3]{\left(1+lnx\right)^4}+C\)

f/ \(I=\int cosx.sin^3xdx\)

Đặt \(u=sinx\Rightarrow du=cosxdx\)

\(\Rightarrow I=\int u^3du=\frac{1}{4}u^4+C=\frac{1}{4}sin^4x+C\)

Từ phần này trở đi mới bắt đầu xài nguyên hàm từng phần:

g/ \(I=\int\left(x^2+2x-1\right)e^xdx\)

Đặt \(\left\{{}\begin{matrix}u=x^2+2x-1\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\left(2x+2\right)dx\\v=e^x\end{matrix}\right.\)

\(\Rightarrow I=\left(x^2+2x-1\right)e^x-\int\left(2x+2\right)e^xdx\)

Xét \(J=\int\left(2x+2\right)e^xdx\)

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

\(\Rightarrow J=\left(2x+2\right)e^x-\int2e^xdx=\left(2x+2\right)e^x-2e^x+C=2x.e^x+C\)

\(\Rightarrow I=\left(x^2+2x-1\right)e^x-2x.e^x+C=\left(x^2-1\right)e^x+C\)

1) Đặt \(t=1+\sqrt{x-1}\Leftrightarrow x=\left(t-1\right)^2+1\forall t\ge1\Rightarrow dx=d\left(t-1\right)^2=2dt\)

\(\Rightarrow I_1=\int\frac{\left(t-1\right)^2+1}{t}\cdot2dt=2\int\frac{t^2-2t+2}{t}dt=2\int\left(t-2+\frac{2}{t}\right)dt\\ =t^2-4t+4lnt+C\)

Thay x vào ta có...

2) \(I_2=\int\frac{2sinx\cdot cosx}{cos^3x-\left(1-cos^2x\right)-1}dx=\int\frac{-2cosx\cdot d\left(cosx\right)}{cos^3x+cos^2x-2}=\int\frac{-2t\cdot dt}{t^3+t-2}\)

\(I_2=\int\frac{-2t}{\left(t-1\right)\left(t^2+2t+2\right)}dt=-\frac{2}{5}\int\frac{dt}{t-1}+\frac{1}{5}\int\frac{2t+2}{t^2+2t+2}dt-\frac{6}{5}\int\frac{dt}{\left(t+1\right)^2+1}\)

Ta có:

\(\int\frac{2t+2}{t^2+2t+2}dt=\int\frac{d\left(t^2+2t+2\right)}{t^2+2t+2}=ln\left(t^2+2t+2\right)+C\)

\(\int\frac{dt}{\left(t+1\right)^2+1}=\int\frac{\frac{1}{cos^2m}}{tan^2m+1}dm=\int dm=m+C=arctan\left(t+1\right)+C\)

Thay x vào, ta có....

3)

\(\frac{1}{\left(1+\sqrt{x}\right)+\sqrt{x+1}}=\frac{\left(1+\sqrt{x}\right)-\sqrt{x+1}}{\left[\left(1+\sqrt{x}\right)-\sqrt{x+1}\right]\cdot\left[\left(1+\sqrt{x}\right)+\sqrt{x+1}\right]}\\ =\frac{\left(1+\sqrt{x}\right)-\sqrt{x+1}}{2\sqrt{x}}=\frac{1}{2\sqrt{x}}+\frac{1}{2}+\frac{\sqrt{x+1}}{2\sqrt{x}}\)

\(I_3=\int\left(\frac{1}{2\sqrt{x}}+\frac{1}{2}+\frac{\sqrt{x+1}}{2\sqrt{x}}\right)dx=\sqrt{x}+\frac{x}{2}+\int\sqrt{\frac{x+1}{x}}\cdot\frac{dx}{2}\)

Xét \(\int\sqrt{\frac{x+1}{x}}\cdot\frac{dx}{2}\)

Đặt \(x=tan^2t\Leftrightarrow dx=\frac{2tant}{cos^2t}\cdot dt\)

\(\Rightarrow\int\sqrt{\frac{x+1}{x}}\cdot\frac{dx}{2}=\int\sqrt{\frac{tan^2t+1}{tan^2t}}\cdot\frac{tant}{cos^2t}dt\\ =\int\frac{1}{sin^2t}\cdot\frac{sint}{cos^3t}dt=\int\frac{d\left(cost\right)}{cos^3t\left(1-cos^2t\right)}=...\)

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

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

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