\(I=\int\left(tanx+tan^3x\right)dx=\int tanx\left(1+tan^2x\right)dx\)

\(I=\int tanx\dfrac{1}{cos^2x}dx\)

Đặt \(tanx=t\Rightarrow\dfrac{dx}{cos^2x}=dt\)

\(\Rightarrow I=\int t.dt=\dfrac{t^2}{2}+C=\dfrac{tan^2x}{2}+C\)

a.

\(I=\int\frac{\frac{1}{2}\left(2x-2\right)+7}{\sqrt{x^2-2x+10}}dx=\frac{1}{2}\int\frac{2x-2}{\sqrt{x^2-2x+10}}dx+7\int\frac{1}{\sqrt{x^2-2x+10}}dx=\frac{1}{2}I_1+7I_2\)

Xét \(I_1=\int\frac{2x-2}{\sqrt{x^2-2x+10}}dx=\int\frac{d\left(x^2-2x+10\right)}{\sqrt{x^2-2x+10}}=2\sqrt{x^2-2x+10}+C_1\)

Xét \(I_2=\int\frac{dx}{\sqrt{x^2-2x+10}}=\int\frac{dx}{\sqrt{\left(x-1\right)^2+9}}\)

Đặt

\(u=x-1+\sqrt{\left(x-1\right)^2+10}\Rightarrow du=\left(1+\frac{\left(x-1\right)}{\sqrt{\left(x-1\right)^2+10}}\right)dx=\frac{x-1+\sqrt{\left(x-1\right)^2+10}}{\sqrt{\left(x-1\right)^2+10}}dx\)

\(\Rightarrow du=\frac{u}{\sqrt{\left(x-1\right)^2+10}}dx\Rightarrow\frac{dx}{\sqrt{\left(x-1\right)^2+10}}=\frac{du}{u}\)

\(\Rightarrow I_2=\int\frac{du}{u}=ln\left|u\right|+C_2=ln\left|x-1+\sqrt{x^2-2x+10}\right|+C_2\)

\(\Rightarrow I=\sqrt{x^2-2x+10}+7ln\left|x-1+\sqrt{x^2-2x+10}\right|+C\)

2.

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

Xét \(I_1=\int\frac{2x+2}{\sqrt{3-2x-x^2}}dx=-\int\frac{d\left(3-2x-x^2\right)}{\sqrt{3-2x-x^2}}=-2\sqrt{3-2x-x^2}+C_1\)

Xét \(I_2=\int\frac{1}{\sqrt{3-2x-x^2}}dx=\int\frac{1}{\sqrt{4-\left(x+1\right)^2}}dx\)

Đặt \(x+1=2sinu\Rightarrow dx=2cosu.du\)

\(\Rightarrow I_2=\int\frac{2cosu.du}{2.cosu}=\int du=u+C_2=arcsin\left(\frac{x+1}{2}\right)+C_2\)

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

c/

\(I=\int\frac{1-\sqrt{x}}{\sqrt{1-x}}dx\)

Đặt \(\sqrt{x}=sint\Rightarrow x=sin^2t\Rightarrow dx=2sint.cost.dt\)

\(\Rightarrow I=\int\frac{2sint.cost\left(1-sint\right)}{\sqrt{1-sin^2t}}dt=\int\frac{2sint.cost\left(1-sint\right)}{cost}dt=\int\left(2sint-2sin^2t\right)dt\)

\(=\int\left(2sint+cos2t-1\right)dt=-2cost+\frac{1}{2}sin2t-t+C\)

\(=-2\sqrt{1-sin^2t}+\frac{1}{2}sint\sqrt{1-sin^2t}-t+C\)

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

Đặt \(u=\ln^3x\rightarrow du=3\ln^2x\frac{dx}{x},dv=dx\rightarrow v=x\)

Do đó : \(I=x\ln^3x|^e_1-3\int\limits^3_1\ln^2xdx=e-3J\left(1\right)\)

Tính \(J=\int\limits^e_1\ln^2xdx\)

Đặt \(u_1=\ln^2x\rightarrow du_1=\frac{2\ln x}{x}dx,dv_1=dx\rightarrow v_1=x\)

Do vậy, \(J=x\ln^2x|^e_1-2\int\limits^e_1\ln xdx=e-2\left(x\ln x|^e_1-\int\limits^e_1dx\right)=e-2\left(x\ln x-x\right)|^e_1=e-2\)

Thay vào (1) ta có : \(I=e-3\left(e-2\right)=6-2e\)

a) \(I_1=\int\frac{dx}{2\sin x\cos x}=\frac{1}{2}\int\frac{\cos x}{\sin x}.\frac{dx}{\cos^2x}\)

Đặt \(\tan x=t\)

        \(=\frac{1}{2}\int\frac{dt}{t}=\frac{1}{2}\ln\left|t\right|+C=\frac{1}{2}\ln\left|\tan x\right|+C\) 

b) \(I_2=\int\frac{\sin^4x}{\cos^4x}.\frac{1}{\cos^2x}.\frac{dx}{\cos^2x}\) 

Đặt \(t=\tan x\)

         \(=\int t^4\left(1+t^2\right)dt\)

         \(=\int t^4dt+\int t^6dt=\frac{t^5}{5}+\frac{t^7}{7}+C\)

         \(=\frac{\tan^5x}{5}+\frac{\tan^7x}{7}+C\)

c) \(I_3=\int\tan^3xdx\)  đặt \(t=\tan x\) 

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

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

       \(=\frac{1}{2}\tan^2x+\ln\left|\cos x\right|+C\)

d) \(\int\frac{dx}{\sin^4x}=\int\frac{1}{\sin^2x}.\frac{1}{\sin^2x}dx=-\int\left(1+\cot^2x\right)d\left(\cot x\right)\)

                                               \(=-\cot x-\frac{1}{3}\cot^3x+C\)

Câu 1)

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

\(\Rightarrow I=x\ln ^3x-3\int \ln^2xdx\)

Tiếp tục nguyên hàm từng phần cho \(\int \ln ^2xdx\) như trên, ta suy ra:

\(\int\ln ^2xdx=x\ln^2x-2\int \ln x dx\).

Tiếp tục nguyên hàm từng phần cho \(\int \ln xdx\Rightarrow \int \ln xdx=x\ln x-x+c\)

Do đó mà \(I=x\ln ^3x-3(x\ln^2x-2x\ln x+2x)+c\)

\(\Leftrightarrow I=x\ln^3x-3x\ln^2x+6x\ln x-6x+c\)

Câu 2)

\(I=\int ^{1}_{0}(x+\sin ^2x)\cos x dx=\int ^{1}_{0}x\cos xdx+\int ^{1}_{0}\sin^2x\cos xdx\)

Đặt \(\left\{\begin{matrix} u=x\\ dv=\cos xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\sin x\end{matrix}\right.\Rightarrow \int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x+c\)

\(\Rightarrow \int ^{1}_{0} x\cos xdx=\sin 1+\cos 1-1\)

Còn \(\int ^{1}_{0}\sin^2x\cos xdx=\int ^{1}_{0}\sin ^2xd(\sin x)=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\sin ^3x}{3}=\frac{\sin^31}{3}\)

\(\Rightarrow I=-1+\sin 1+\cos 1+\frac{\sin ^3 1}{3}\approx 0,0173\)

Câu 3:

Đối với \(\int xe^{2x}dx\)

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

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

Đối với \(\int x\sqrt[3]{x+1}dx=\int \sqrt[3]{(x+1)^4}dx-\int \sqrt{x+1}dx=\frac{3(x+1)^\frac{7}{3}}{7}-\frac{3}{4}(x+1)^{\frac{4}{3}}+c\)

\(\Rightarrow \int x\sqrt[3]{x+1}dx=\frac{3(x+1)^{\frac{4}{3}}(4x-3)}{28}\)

Do đó mà \(\int x(e^{2x}-\sqrt[3]{x+1})dx=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}+\frac{3(x+1)^{\frac{4}{3}}(4x-3)}{28}+c\)

4 câu 1,3,4,5 giống nhau, mình làm 1 câu và bạn dựa vào đó tự xử lý mấy câu còn lại nhé

1/ \(I=\int sin2x.e^{3x}dx\) \(\Rightarrow\left\{{}\begin{matrix}u=sin2x\\dv=e^{3x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2cos2x.dx\\v=\dfrac{1}{3}e^{3x}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{1}{3}sin2x.e^{3x}-\dfrac{2}{3}\int cos2x.e^{3x}dx=\dfrac{1}{3}sin2x.e^{3x}-\dfrac{2}{3}I_1\)

Xét \(I_1=\int cos2x.e^{3x}dx\) \(\Rightarrow\left\{{}\begin{matrix}u=cos2x\\dv=e^{3x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=-2sin2xdx\\v=\dfrac{1}{3}e^{3x}\end{matrix}\right.\)

\(\Rightarrow I_1=\dfrac{1}{3}cos2x.e^{3x}+\dfrac{2}{3}\int sin2x.e^{3x}dx=\dfrac{1}{3}cos2x.e^{3x}+\dfrac{2}{3}I\)

\(\Rightarrow I=\dfrac{1}{3}sin2x.e^{3x}-\dfrac{2}{3}\left(\dfrac{1}{3}cos2x.e^{3x}+\dfrac{2}{3}I\right)\)

\(\Rightarrow\dfrac{13}{9}I=\dfrac{1}{9}e^{3x}\left(3sin2x-2cos2x\right)\)

\(\Rightarrow I=\dfrac{1}{13}e^{3x}\left(3sin2x-2cos2x\right)+C\)

3/ \(\int e^x\left(\dfrac{1+cos2x}{2}\right)dx=\dfrac{1}{2}\int e^xdx+\dfrac{1}{2}\int cos2x.e^xdx=\dfrac{e^x}{2}+\dfrac{1}{2}I_1\)

\(I_1\) có cách tính y hệt như bài 1, bạn nguyên hàm từng phần 2 lần là xong

4/ Cũng hạ bậc tương tự câu trên và xử lý

5/ \(I=\int e^{-x}\left(\dfrac{cos3x+3cosx}{4}\right)dx=\dfrac{1}{4}\int e^{-x}\left(cos3x+3cosx\right)dx\)

\(\Rightarrow I=\dfrac{1}{4}\int e^{-x}cos3x.dx+\dfrac{3}{4}\int e^{-x}cosx.dx=I_1+I_2\)

Dùng phương pháp tương tự bài 1, lần lượt tính \(I_1\) và \(I_2\) rồi cộng vào

2/\(I=\int\dfrac{x^4}{\left(x^2-1\right)^2}dx=\int\left(1+\dfrac{2x^2-1}{\left(x^2-1\right)^2}\right)dx=\int\left(1+\dfrac{2}{x^2-1}+\dfrac{1}{\left(x^2-1\right)^2}\right)dx\)

\(=\int\left(1+\dfrac{1}{x-1}-\dfrac{1}{x+1}+\dfrac{1}{4}\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)^2\right)dx\)

\(=\int\left(1+\dfrac{1}{x-1}-\dfrac{1}{x+1}+\dfrac{1}{4}\left(\dfrac{1}{\left(x-1\right)^2}+\dfrac{1}{\left(x+1\right)^2}+\dfrac{1}{x+1}-\dfrac{1}{x-1}\right)\right)dx\)

\(=\int\left(1+\dfrac{3}{4}\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)+\dfrac{1}{4}\dfrac{1}{\left(x+1\right)^2}+\dfrac{1}{4}\dfrac{1}{\left(x-1\right)^2}\right)dx\)

\(=x+\dfrac{3}{4}ln\left|\dfrac{x-1}{x+1}\right|-\dfrac{1}{4\left(x+1\right)}-\dfrac{1}{4\left(x-1\right)}+C\)

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

a/ Đặt \(lnx=t\Rightarrow\frac{dx}{x}=dt\)

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

\(=\frac{ln^3x}{6}+\frac{lnx}{2}+C\)

b/ \(I=\int sin^2x.cos^2x.cosxdx=\int sin^2x\left(1-sin^2x\right)cosxdx\)

Đặt \(sinx=t\Rightarrow cosxdx=dt\)

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

\(=\frac{1}{3}sin^3x-\frac{1}{5}sin^5x+C\)

c/ \(I=\int x^4\sqrt{x^2+1}xdx\)

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

\(\Rightarrow I=\int\left(t^2-1\right)^2.t.tdt=\int\left(t^4-2t^2+1\right)t^2dt\)

\(=\int\left(t^6-2t^4+t^2\right)dt=\frac{1}{7}t^7-\frac{2}{5}t^5+\frac{1}{3}t^3+C\)

\(=\frac{1}{7}\sqrt{\left(x^2+1\right)^7}-\frac{2}{5}\sqrt{\left(x^2+1\right)^5}+\frac{1}{3}\sqrt{\left(x^2+1\right)^3}+C\)

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

\(\Rightarrow I=\int\frac{2\left(t-1\right)dt}{t}=\int\left(2-\frac{2}{t}\right)dt=2t-2lnt+C\)

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

\(a,\int sin2x.cosxdx=\int\dfrac{1}{2}\left[sin3x+sinx\right]dx=\dfrac{1}{2}\int sin3xdx+\dfrac{1}{2}\int sinxdx=\dfrac{-1}{6}cos3x-\dfrac{1}{2}cosx\)

phần a bạn thêm +C vào đáp án nhé
\(i,\int2sinx3x.cos2xdx=2\int\dfrac{1}{2}\left(sin5x+sinx\right)dx=\int sin5xdx+\int sinxdx=-\dfrac{1}{5}cos5x-cosx+C\)

\(g,\int\dfrac{1}{sin^2x.cos^2x}=\int\dfrac{sin^2x+cos^2x}{sin^2x.cos^2x}=\int\dfrac{1}{cos^2x}dx+\int\dfrac{1}{sin^2x}dx=tanx-cotx+C\)

\(I=\int tan^2x\left(\dfrac{1}{cos^2x}-1\right)dx=\int tan^2x.\dfrac{1}{cos^2x}dx-\int tan^2xdx\)

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