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 \(\sqrt{2x-5}=t\) 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.

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