a) Đặt \(1+\ln x=t\)  khi đó \(\frac{dx}{x}=dt\)  và do đó 

\(I_1=\int\sqrt{t}dt=\frac{2}{3}t^{\frac{3}{2}}+C=\frac{2}{3}\sqrt{\left(1+\ln x\right)^3}+C\)

b) Đặt \(\sqrt[4]{e^x+1}=t\)  khi đó \(e^x+1=t^4\Rightarrow e^x=t^4-1\) và \(e^xdx=4t^3dt\)  , \(e^{2x}dx=e^x.e^xdx=\left(t^4-1\right)4t^3dt\) 

Do đó :

\(I_2=4\int\frac{t^3\left(t^4-1\right)}{t}dt=4\int\left(t^6-t^2\right)dt=4\left[\frac{t^7}{7}-\frac{t^3}{3}\right]+C\)

    \(=4\left[\frac{1}{7}\sqrt[4]{\left(e^x+1\right)^7}-\frac{1}{3}\sqrt[4]{\left(e^x+1\right)^3}\right]+C\)

c) Lưu ý rằng \(x^2dx=\frac{1}{3}d\left(x^3+C\right)\) do đó :

\(I_3=\int x^2e^{x^{3+6}dx}=\frac{1}{3}\int e^{x^{3+6}}d\left(x^3+6\right)=\frac{1}{3}e^{x^{3+6}}+C\)

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.

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

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

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

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

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

a: \(I_1 = \int \left( \tan(x) - \ln^{15}(\cos(x)) \right) dx\)

=>\(I_1 = \int \tan(x) \, dx - \int \ln^{15}(\cos(x)) \, dx\)

\(A=\int\tan(x)\,dx\)

\(=\int\frac{\sin(x)}{\cos(x)}\,dx\)

\(=-\int\frac{d(\cos(x))}{\cos(x)}=-\ln\vert{}\cos(x)\vert{}\)

\(B = \int \ln^{15}(\cos(x)) \, dx\)

Đặt \(u=\ln(\cos(x))\)

\(\implies du=\frac{-\sin(x)}{\cos(x)}dx=-\tan(x)dx\)

\(\int \tan(x) \ln^{15}(\cos(x)) \, dx = -\int u^{15} \, du = -\frac{u^{16}}{16} + C = -\frac{\ln^{16}(\cos(x))}{16} + C\)

Do đó: \(I_1 = -\ln\vert{}\cos(x)\vert{} - \int \ln^{15}(\cos(x)) \, dx + C\)

b: \(I_2 = \int \frac{x^4 + x^2 + 1}{2x^3 + 5x^2 - 7} \, dx\)

Ta có: \(x^4 + x^2 + 1 = \left( \frac{x}{2} - \frac{5}{4} \right)(2x^3 + 5x^2 - 7) + \left( \frac{25}{4}x^2 + \frac{7}{2}x - \frac{31}{4} \right)\)

=>\(\frac{x^4 + x^2 + 1}{2x^3 + 5x^2 - 7}=\frac{x}{2}-\frac{5}{4}+\frac{\frac{25}{4}x^2 + \frac{7}{2}x - \frac{31}{4}}{2x^3 + 5x^2 - 7}\)

\(=\frac{x}{2}-\frac{5}{4}+\frac{25x^2 + 14x - 31}{4(2x^3 + 5x^2 - 7)}\)

Đặt \(\frac{25x^2 + 14x - 31}{(x - 1)(2x^2 + 7x + 7)} = \frac{A}{x - 1} + \frac{Bx + C}{2x^2 + 7x + 7}\)

=>\(25x^2 + 14x - 31 = A(2x^2 + 7x + 7) + (Bx + C)(x - 1)\)

=>\(25x^2+14x-31=x^2\left(2A+B\right)+x\left(7A-B+C\right)+7A-C\)

=>\(\begin{cases}2A+B=25\\ 7A-B+C=14\\ 7A-C=-31\end{cases}\Rightarrow\begin{cases}2A+B=25\\ 7A-B+C-7A+C=14+31\\ 7A-C=-31\end{cases}\)

=>2A+B=25 và -B+2C=45 và 7A-C=-31

=>B=25-2A và -25+2A+2C=45 và 7A-C=-31

=>2A+2C=70 và 7A-C=-31 và B=25-2A

=>A+C=35 và 7A-C=-31 và B=25-2A

=>8A=4 và A+C=35 và B=25-2A

=>A=1/2; C=35-1/2=69/2; B=25-2*1/2=24

Do đó: \(\frac{25x^2 + 14x - 31}{4(2x^3 + 5x^2 - 7)} = \frac{1}{8(x - 1)} + \frac{24x + \frac{69}{2}}{4(2x^2 + 7x + 7)} = \frac{1}{8(x - 1)} + \frac{48x + 69}{8(2x^2 + 7x + 7)}\)

48x+69=12(4x+7)-15

=>\(\int \frac{48x + 69}{2x^2 + 7x + 7} dx = 12 \int \frac{4x + 7}{2x^2 + 7x + 7} dx - 15 \int \frac{dx}{2x^2 + 7x + 7}\)

\(= 12 \ln(2x^2 + 7x + 7) - \frac{15}{2} \int \frac{dx}{\left(x + \frac{7}{4}\right)^2 + \frac{7}{16}}\)

\(= 12 \ln(2x^2 + 7x + 7) - \frac{30}{\sqrt{7}} \arctan\left( \frac{4x + 7}{\sqrt{7}} \right)\)

=>\(I_2 = \int \left( \frac{x}{2} - \frac{5}{4} \right) dx + \frac{1}{8} \int \frac{dx}{x - 1} + \frac{1}{8} \int \frac{48x + 69}{2x^2 + 7x + 7} dx\)

\(=\frac{x^2}{4}-\frac{5x}{4}+\frac{1}{8}\ln\vert{}x-1\vert{}+\frac{3}{2}\ln(2x^2+7x+7)-\frac{15}{4\sqrt{7}}\arctan\left(\frac{4x + 7}{\sqrt{7}}\right)+C\)

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 \(\Rightarrow \int t\cos tdt=t\sin t+\cos t+c\)

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

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 \(I=\frac{-1}{2\sqrt{3}}v+c\)

Lời giải:

Đặt \(u=\ln (x+\sqrt{x^2+1}); dv=\frac{1}{\sqrt{x^2+1}}dx\)

\(\Rightarrow du=\frac{dx}{\sqrt{x^2+1}}; v=\int \frac{x}{\sqrt{x^2+1}}dx=\frac{1}{2}\int \frac{d(x^2+1)}{\sqrt{x^2+1}}=\sqrt{x^2+1}\)

\(\Rightarrow \int \frac{x\ln (x+\sqrt{x^2+1})}{\sqrt{x^2+1}}dx=\int udv=uv-vdu=\sqrt{x^2+1}\ln (x+\sqrt{x^2+1})-\int dx\)

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