\(a=\int\dfrac{1}{2tan^2x+5tanx+2}.\dfrac{dx}{cos^2x}\)

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

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

\(=\dfrac{1}{3}ln\left|\dfrac{2t+1}{2t+4}\right|+C=\dfrac{1}{3}ln\left|\dfrac{2tanx+1}{2tanx+4}\right|+C\)

Câu b hoàn toàn tương tự

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

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

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

1.

\(I=\int\dfrac{cot^2x}{sin^6x}dx=\int\dfrac{cot^2x}{sin^4x}.\dfrac{1}{sin^2x}=\int cot^2x\left(1+cot^2x\right)^2.\dfrac{1}{sin^2x}dx\)

Đặt \(u=cotx\Rightarrow du=-\dfrac{1}{sin^2x}dx\)

\(I=-\int u^2\left(1+u^2\right)^2du=-\int\left(u^6+2u^4+u^2\right)du\)

\(=-\dfrac{1}{7}u^7+\dfrac{2}{5}u^5+\dfrac{1}{3}u^3+C\)

\(=-\dfrac{1}{7}cot^7x+\dfrac{2}{5}cot^5x+\dfrac{1}{3}cot^3x+C\)

2.

\(I=\int\left(e^{sinx}+cosx\right).cosxdx=\int e^{sinx}.cosxdx+\int cos^2xdx\)

\(=\int e^{sinx}.d\left(sinx\right)+\dfrac{1}{2}\int\left(1+cos2x\right)dx\)

\(=e^{sinx}+\dfrac{1}{2}x+\dfrac{1}{4}sin2x+C\)

\(\int\left(3x^2-2x-4\right)dx=x^3-x^2-4x+C\)

\(\int\left(sin3x-cos4x\right)dx=-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+C\)

\(\int\left(e^{-3x}-4^x\right)dx=-\dfrac{1}{3}e^{-3x}-\dfrac{4^x}{ln4}+C\)

d. \(I=\int lnxdx\)

Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=x\end{matrix}\right.\)

\(\Rightarrow u=x.lnx-\int dx=x.lnx-x+C\)

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

\(\Rightarrow I=x.e^x-\int e^xdx=x.e^x-e^x+C\)

f.

Đặt \(\left\{{}\begin{matrix}u=x+1\\dv=sinxdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=-cosx\end{matrix}\right.\)

\(\Rightarrow I=-\left(x+1\right)cosx+\int cosxdx=-\left(x+1\right)cosx+sinx+C\)

g.

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

\(\Rightarrow I=\dfrac{1}{2}x^2.lnx-\dfrac{1}{2}\int xdx=\dfrac{1}{2}x^2.lnx-\dfrac{1}{4}x^2+C\)

\(=-\dfrac{1}{2}\int\dfrac{d\left(4-2x\right)}{\left(4-2x\right)^5}=\dfrac{1}{8.\left(4-2x\right)^4}+C\)

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

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