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

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

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

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

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

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

Câu 1)

Ta có \(I=\int ^{1}_{0}\frac{dx}{\sqrt{3+2x-x^2}}=\int ^{1}_{0}\frac{dx}{4-(x-1)^2}\).

Đặt \(x-1=2\cos t\Rightarrow \sqrt{4-(x-1)^2}=\sqrt{4-4\cos^2t}=2|\sin t|\)

Khi đó:

\(I=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{d(2\cos t+1)}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{2\sin tdt}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}dt=\left.\begin{matrix} \frac{2\pi}{3}\\ \frac{\pi}{2}\end{matrix}\right|t=\frac{\pi}{6}\)

Câu 3)

\(K=\int ^{3}_{2}\ln (x^3-3x+2)dx=\int ^{3}_{2}\ln [(x+2)(x-1)^2]dx\)

\(=\int ^{3}_{2}\ln (x+2)d(x+2)+2\int ^{3}_{2}\ln (x-1)d(x-1)\)

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

\(\Rightarrow K=\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x+2)[\ln (x+2)-1]+2\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x-1)[\ln (x-1)-1]\)

\(=5\ln 5-4\ln 4-1+4\ln 2-2=5\ln 5-4\ln 2-3\)

Bài 2)

\(J=\int ^{1}_{0}x\ln (2x+1)dx\). Đặt \(\left\{\begin{matrix} u=\ln (2x+1)\\ dv=xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2dx}{2x+1}\\ v=\frac{x^2}{2}\end{matrix}\right.\)

Khi đó:

\(J=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2\ln (2x+1)}{2}-\int ^{1}_{0}\frac{x^2}{2x+1}dx\)\(=\frac{\ln 3}{2}-\frac{1}{4}\int ^{1}_{0}(2x-1+\frac{1}{2x+1})dx\)

\(=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2-x}{4}-\frac{1}{8}\int ^{1}_{0}\frac{d(2x+1)}{2x+1}=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\ln (2x+1)}{8}\)

\(=\frac{\ln 3}{2}-\frac{\ln 3}{8}=\frac{3\ln 3}{8}\)

Câu 5)

\(J=\underbrace{\int ^{3}_{1}\frac{3dx}{(x+1)^2}}_{A}+\underbrace{\int ^{3}_{1}\frac{\ln xdx}{(x+1)^2}}_{B}\)

Ta có: \(A=\int ^{3}_{1}\frac{3d(x+1)}{(x+1)^2}=\left.\begin{matrix} 3\\ 1\end{matrix}\right|\frac{-3}{x+1}=\frac{3}{4}\)

\(B=\int ^{3}_{1}\frac{\ln xdx}{(x+1)^2}=\left.\begin{matrix} 3\\ 1\end{matrix}\right|\frac{-\ln x}{x+1}+\int ^{3}_{1}\frac{dx}{x(x+1)}=\frac{-\ln 3}{4}+\left.\begin{matrix} 3\\ 1\end{matrix}\right|(\ln |x|-\ln|x+1|)\)

\(B=\frac{-\ln 3}{4}+(\ln 3-\ln 4)+\ln 2=\frac{3}{4}\ln 3-\ln 2\)

a) Đặt \(\sqrt{2x-5}=t\) khi đó \(x=\frac{t^2+5}{2}\) , \(dx=tdt\)

Do vậy \(I_1=\int\frac{\frac{1}{4}\left(t^2+5\right)^2+3}{t^3}dt=\frac{1}{4}\int\frac{\left(t^4+10t^2+37\right)t}{t^3}dt\)

                \(=\frac{1}{4}\int\left(t^2+10+\frac{37}{t^2}\right)dt=\frac{1}{4}\left(\frac{t^3}{3}+10t-\frac{37}{t}\right)+C\)

Trở về biến x, thu được :

\(I_1=\frac{1}{12}\sqrt{\left(2x-5\right)^3}+\frac{5}{2}\sqrt{2x-5}-\frac{37}{4\sqrt{2x-5}}+C\)

b) \(I_2=\frac{1}{3}\int\frac{d\left(\ln\left(3x-1\right)\right)}{\ln\left(3x-1\right)}=\frac{1}{3}\ln\left|\ln\left(3x-1\right)\right|+C\)

c) \(I_3=\int\frac{1+\frac{1}{x^2}}{\sqrt{x^2-7+\frac{1}{x^2}}}dx=\int\frac{d\left(x-\frac{1}{x}\right)}{\sqrt{\left(x-\frac{1}{2}\right)^2-5}}\)

Đặt \(x-\frac{1}{x}=t\)

\(\Rightarrow\) \(I_3=\int\frac{dt}{\sqrt{t^2-5}}=\ln\left|t+\sqrt{t^2-5}\right|+C\)

                           \(=\ln\left|x-\frac{1}{x}+\sqrt{x^2-7+\frac{1}{x^2}}\right|+C\)

Chịu thôi khó quá.

1/ \(\int\limits^e_1\left(x+\frac{1}{x}+\frac{1}{x^2}\right)dx=\left(\frac{x^2}{2}+lnx-\frac{1}{x}\right)|^e_1=\frac{e^2}{2}-\frac{1}{e}+\frac{3}{2}\)

2/ \(\int\limits^2_1\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)dx=\int\limits^2_1\left(x\sqrt{x}+1\right)dx=\int\limits^2_1\left(x^{\frac{3}{2}}+1\right)dx\)

\(=\left(\frac{2}{5}.x^{\frac{5}{2}}+x\right)|^2_1=\frac{8\sqrt{2}-7}{5}\)

3/

\(\int\limits^2_1\frac{2x^3-4x+5}{x}dx=\int\limits^2_1\left(2x^2-4+\frac{5}{x}\right)dx=\left(\frac{2}{3}x^3-4x+5lnx\right)|^2_1=\frac{2}{3}+5ln2\)

4/ \(\int\limits^2_1x^2\left(3x-1\right)\frac{2}{x}dx=\int\limits^2_1\left(6x^2-2x\right)dx=\left(2x^3-x^2\right)|^2_1=11\)

a) Dùng phương pháp hữu tỉ hóa "Nếu \(f\left(x\right)=R\left(e^x\right)\Rightarrow t=e^x\)"  ta có \(e^x=t\Rightarrow x=\ln t,dx=\frac{dt}{t}\)

Khi đó \(I_1=\int\frac{t^3}{t+2}.\frac{dt}{t}=\int\frac{t^2}{t+2}dt=\int\left(t-2+\frac{4}{t+2}\right)dt\)

                \(=\frac{1}{2}t^2-2t+4\ln\left(t+2\right)+C=\frac{1}{2}e^{2x}-2e^x+4\ln\left(e^x+2\right)+C\)

b) Hàm dưới dấu nguyên hàm

\(f\left(x\right)=\frac{\sqrt{x}}{x+\sqrt[3]{x^2}}=R\left(x;x^{\frac{1}{2}},x^{\frac{2}{3}}\right)\)

q=BCNN(2;3)=6

Ta thực hiện phép hữu tỉ hóa theo :

"Nếu \(f\left(x\right)=R\left(x:\left(ã+b\right);\left(ax+b\right)^{r2},....\right),r_k=\frac{P_k}{q_k}\in Q,k=1,2,...,m\Rightarrow t=\left(ax+b\right)^{\frac{1}{q}}\),q=BCNN \(\left(q_1,q_2,...,q_m\right)\)"

=> \(t=x^{\frac{1}{6}}\Rightarrow x=t^{6,}dx=6t^5dt\)

Khi đó nguyên hàm đã cho trở thành :

\(I_2=\int\frac{t^3}{t^6-t^4}6t^{5dt}=\int\frac{6t^4}{t^2-1}dt=6\int\left(t^2+1+\frac{1}{t^2-1}\right)dt\)

     \(=6\int\left(t^2+1\right)dt+2\int\frac{dt}{\left(t-1\right)\left(t+1\right)}=2t^3+6t+3\int\frac{dt}{t-1}-3\int\frac{dt}{t+1}\)

     \(=2t^2+6t+3\ln\left|t-1\right|-3\ln\left|t+1\right|+C=2\sqrt{x}+6\sqrt[6]{x}+3\ln\left|\frac{\sqrt[6]{x-1}}{\sqrt[6]{x+1}}\right|+C\)

c) Hàm dưới dấu nguyên hàm có dạng :

\(f\left(x\right)=R\left(x;\left(\frac{x+1}{x-1}\right)^{\frac{2}{3}};\left(\frac{x+1}{x-1}\right)^{\frac{5}{6}}\right)\)

q=BCNN (3;6)=6

Ta thực hiện phép hữu tỉ hóa được

\(t=\left(\frac{x+1}{x-1}\right)^{\frac{1}{6}}\Rightarrow x=\frac{t^6+1}{t^6-1},dx=\frac{-12t^5}{\left(t^6-1\right)^2}dt\)

Khi đó hàm dưới dấu nguyên hàm trở thành

\(R\left(t\right)=\frac{1}{\left(\frac{t^6+1}{t^6-1}\right)^2-1}\left[t^4-t^5\right]=\frac{\left(t^6-1\right)^2}{4t^6}\left(t^4-t^5\right)\)

Do đó :

\(I_3=\int\frac{\left(t^6-1\right)^2}{4t^6}\left(t^4-t^5\right).\frac{-12t^5}{\left(t^6-1\right)}dt=3\int\left(t^4-t^3\right)dt\)

    \(=\frac{5}{3}t^5-\frac{3}{4}t^4+C=\frac{3}{5}\sqrt[6]{\left(\frac{x+1}{x-1}\right)^5}-\frac{3}{4}\sqrt[3]{\left(\frac{x+1}{x-1}\right)^2}+C\)

Câu 1:

\(\int\limits^3_0\left(f'\left(x\right)+1\right)\sqrt{x+1}dx=\int\limits^3_0f'\left(x\right)\sqrt{x+1}dx+\int\limits^3_0\sqrt{x+1}dx\)

\(=\int\limits^3_0f'\left(x\right)\sqrt{x+1}dx+\frac{14}{3}=\frac{302}{15}\Rightarrow\int\limits^1_0f'\left(x\right)\sqrt{x+1}dx=\frac{232}{15}\)

Ta có:

\(I=\int\limits^3_0\frac{f\left(x\right)dx}{\sqrt{x+1}}\)

Đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=\frac{dx}{\sqrt{x+1}}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=2\sqrt{x+1}\end{matrix}\right.\)

\(\Rightarrow I=2f\left(x\right)\sqrt{x+1}|^3_0-2\int\limits^3_0f'\left(x\right)\sqrt{x+1}dx\)

\(=4f\left(3\right)-2f\left(0\right)-2.\frac{232}{15}\)

\(=2\left(2f\left(3\right)-f\left(0\right)\right)-\frac{464}{15}=36-\frac{464}{15}=\frac{76}{15}\)

Câu 2:

\(I_1=\int\limits^3_1\frac{xf'\left(x\right)}{x+1}dx=0\)

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

\(\Rightarrow I_1=\frac{xf\left(x\right)}{x+1}|^3_1-\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}=\frac{3.3}{3+1}-\frac{1.3}{1+1}-\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}dx=\frac{3}{4}-\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}dx=0\)

\(\Rightarrow\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}dx=\frac{3}{4}\)

Ta có:

\(I=\int\limits^3_1\frac{f\left(x\right)+lnx}{\left(x+1\right)^2}dx=\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}dx+\int\limits^3_1\frac{lnx}{\left(x+1\right)^2}dx=\frac{3}{4}+I_2\)

Xét \(I_2=\int\limits^3_1\frac{lnx}{\left(x+1\right)^2}dx\Rightarrow\) đặt \(\left\{{}\begin{matrix}u=lnx\\dv=\frac{1}{\left(x+1\right)^2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{dx}{x}\\v=\frac{-1}{x+1}\end{matrix}\right.\)

\(\Rightarrow I_2=\frac{-lnx}{x+1}|^3_1+\int\limits^3_1\frac{dx}{x\left(x+1\right)}=-\frac{1}{4}ln3+\int\limits^1_0\left(\frac{1}{x}-\frac{1}{x+1}\right)dx\)

\(=-\frac{1}{4}ln3+ln\left(\frac{x}{x+1}\right)|^3_1=-\frac{1}{4}ln3+ln\frac{3}{4}-ln\frac{1}{2}=\frac{3}{4}ln3-ln2\)

\(\Rightarrow I=\frac{3}{4}+\frac{3}{4}ln3-ln2\)

lm jup mk di m.n