a) \(\int\dfrac{2dx}{x^2-5x}=\int\left(\dfrac{-2}{5x}+\dfrac{2}{5\left(x-5\right)}\right)dx=-\dfrac{2}{5}ln\left|x\right|+\dfrac{2}{5}ln\left|x-5\right|+C\)

\(\Rightarrow A=-\dfrac{2}{5};B=\dfrac{2}{5}\Rightarrow2A-3B=-2\)

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

\(\Rightarrow A=\dfrac{1}{3};B=\dfrac{1}{2};E=-2\Rightarrow A-B+E=-\dfrac{13}{6}\)

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

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

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

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

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

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

Đặt \(\frac{lnx}{x}=t\Rightarrow\left(\frac{1-lnx}{x^2}\right)dx=dt\)

\(\Rightarrow I=\int\limits^{\frac{1}{e}}_0\frac{dt}{\left(1+t\right)^2}=-\frac{1}{1+t}|^{\frac{1}{e}}_0=\frac{1}{e+1}\)

\(\Rightarrow a=b=1\Rightarrow a^2+b^2=2\)

Câu 1: \(I = \int_{0}^{1} \frac{x+1}{x^2+2x+5} \, dx\)

Đặt \(u=x^2+2x+5\)

=>du=(2x+2)dx=2(x+1)dx

=>\(\left(x+1\right)\cdot\left(dx\right)=\frac{du}{2}\)

Khi x=0 thì u=5

Khi x=1 thì u=8

\(I = \int_{5}^{8} \frac{\frac{du}{2}}{u} = \frac{1}{2} \ln\vert{}u\vert{} \bigg\vert{}_{5}^{8} = \frac{1}{2} (\ln 8 - \ln 5) = \frac{1}{2} \ln \left(\frac{8}{5}\right)\)

=>a=2; b=8; c=5

P=a+b=10

Câu 2:

\(2x^2-3x-5\)

\(=2x^2-5x+2x-5=\left(2x-5\right)\left(x+1\right)\)

Đặt \(\frac{3x+4}{2x^2-3x-5}=\frac{3x+4}{\left(2x-5\right)\left(x+1\right)}=\frac{A}{2x-5}+\frac{B}{x+1}\)

=>\(\frac{3x+4}{\left(2x-5\right)\left(x+1\right)}=\frac{A\left(x+1\right)+B\left(2x-5\right)}{\left(2x-5\right)\left(x+1\right)}\)

=>A(x+1)+B(2x-5)=3x+4

=>x(A+2B)+A-5B=3x+4

=>A+2B=3 và A-5B=4

=>A+2B-A+5B=3-4 và A+2B=3

=>7B=-1 và A=3-2B

=>\(B=-\frac17;A=3-2\cdot\frac{-1}{7}=3+\frac27=\frac{23}{7}\)

\(I = \int_{0}^{1} \left( -\frac{1}{7} \cdot \frac{1}{x+1} + \frac{23}{7} \cdot \frac{1}{2x-5} \right) dx\)

\(=\left[-\frac{1}{7}\ln\vert{}x+1\vert{}+\frac{23}{14}\ln\vert{}2x-5\vert{}\right]_0^1\)

\(=\left(-\frac{1}{7}\ln2+\frac{23}{14}\ln\vert{}-3\vert{}\right)-\left(-\frac{1}{7}\ln1+\frac{23}{14}\ln\vert{}-5\vert{}\right)\)

\(=-\frac{1}{7}\ln2+\frac{23}{14}\ln3-\frac{23}{14}\ln5\)

\(=\frac{23}{14}(\ln3-\ln5)-\frac{1}{7}\ln2=\frac{23}{14}\ln\left(\frac{3}{5}\right)-\frac{1}{7}\ln2\)

=>a=3; b=5; c=2

P=a-b+c

=3-5+2

=5-5

=0

a/ Tích phân này làm sao giải được nhỉ?

b/ Đặt \(\sqrt{x}=t\Rightarrow x=t^2\Rightarrow dx=2t.dt\)

\(I=\int\frac{2t^2.dt}{4-t^4}=\int\left(\frac{1}{2-t^2}-\frac{1}{2+t^2}\right)dt=\frac{1}{2\sqrt{2}}ln\left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|+\frac{1}{\sqrt{2}}arctan\frac{\sqrt{2}}{t}+C\)

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

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

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

\(\Rightarrow I=\int\frac{t^2dt}{t^2-1}=\int\left(1+\frac{1}{t^2-1}\right)dt=t+ln\left|\frac{t-1}{t+1}\right|+C=\sqrt{1+x^2}+ln\left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+C\)

d/ Con nguyên hàm này cũng không tính được, chắc bạn ghi nhầm đề

Đặt \(I = \int_0^1 \frac{\ln(x + \sqrt{1 - x^2})}{x} dx\) và \(J = \int_0^1 \frac{\ln(1 + x)}{x} dx\)

Đặt x=sin t, với t∈[0;pi/2]

=>dx=cost*dt

\(\sin t + \cos t = \sqrt{2} \sin\left(t + \frac{\pi}{4}\right)\)

=>\(\ln(\sin t+\cos t)=\frac{1}{2}\ln2+\ln\left(\sin\left(t+\frac{\pi}{4}\right)\right)\)

\(I = \int_0^{\pi/2} \frac{\ln(\sin t + \cos t)}{\sin t} \cos t \, dt\)

\(=\int_0^{\pi/2}\left(\frac{1}{2}\ln2+\ln\left(\sin\left(t+\frac{\pi}{4}\right)\right)\right)\cot t\,dt\)

\(=\Big[\ln(\sin t)\ln(\sin t+\cos t)\Big]_0^{\pi/2}-\int_0^{\pi/2}\ln(\sin t)\cdot\frac{\cos t - \sin t}{\sin t + \cos t}dt\)

\(=\int_0^{\pi/2}\ln(\sin t)\cdot\frac{\sin t - \cos t}{\sin t + \cos t}dt\)

\(=-\int_0^{\pi/2}\ln(\sin t)\frac{\cos(2t)}{1 + \sin(2t)}dt\)

Đổi biến u=2t

=>\(t=\frac{u}{2};dt=\frac{du}{2}\)

\(I = -\frac{1}{2} \int_0^\pi \ln\left(\sin\frac{u}{2}\right) \frac{\cos u}{1 + \sin u} du\)

\(J = \int_0^1 \frac{\ln(1+x)}{x} dx = \frac{\pi^2}{12}\)

\(\ln\left(x + \sqrt{1-x^2}\right) = x + \sum_{k=1}^\infty \frac{(-1)^k (2k-1)!!}{(2k)!! (2k+1)} x^{2k+1}\)

=>\(I = 1 + \sum_{k=1}^\infty \frac{(-1)^k (2k-1)!!}{(2k)!! (2k+1)^2}\)

\(I = \frac{1}{2} \int_0^{\pi/2} \frac{\ln(\tan t + 1)}{\sin t \cos t} dt - \dots = \frac{\pi^2}{16}\)

\(\frac{I}{J} = \frac{\frac{\pi^2}{16}}{\frac{\pi^2}{12}} = \frac{12}{16} = \frac{3}{4}\)

=>I=3/4J

=>ĐPCM