\(\int tan\left(x\right)-ln^{15}\left(cos\left(x\right)\right)dx\)
\(\int\dfrac{x^4+x^2+1}{2x^3+5x^2-7}dx\)
tính nguyên hàm , ai giúp mình 2 bài này với hoặc 1 bài thôi cũng đc ạ , xin cảm ơn nhiều.
Tính các nguyên hàm sau :
a) \(\int x\left(3-x\right)^5dx\)
b) \(\int\left(2^x-3^x\right)^2dx\)
c) \(\int x\sqrt{2-5x}dx\)
d) \(\int\dfrac{\ln\left(\cos x\right)}{\cos^2x}dx\)
e) \(\int\dfrac{x}{\sin^2x}dx\)
\(\int\dfrac{x+1}{\left(x-2\right)\left(x+3\right)}dx\)
h) \(\int\dfrac{1}{1-\sqrt{x}}dx\)
i) \(\int\sin3x\cos2xdx\)
k) \(\int\dfrac{\sin^3x}{\cos^2x}dx\)
l) \(\int\dfrac{\sin x\cos x}{\sqrt{a^2\sin^2x+b^2\cos^2x}}dx\) (\(a^2\ne b^2\))
1) \(\int ln\frac{\left(1+s\text{inx}\right)^{1+c\text{os}x}}{1+c\text{os}x}dx\)
2) \(\int\left(xlnx\right)^2dx\)
3) \(\int\frac{3xcosx+2}{1+cot^2x}dx\)
4)\(\int\frac{2}{c\text{os}2x-7}dx\)
5)\(\int\frac{1+x\left(2lnx-1\right)}{x\left(x+1\right)^2}dx\)
6) \(\int\frac{1-x^2}{\left(1+x^2\right)^2}dx\)
7)\(\int e^x\frac{1+s\text{inx}}{1+c\text{os}x}dx\)
8) \(\int ln\left(\frac{x+1}{x-1}\right)dx\)
9)\(\int\frac{xln\left(1+x\right)}{\left(1+x^2\right)^2}dx\)
10) \(\int\frac{ln\left(x-1\right)}{\left(x-1\right)^4}dx\)
11)\(\int\frac{x^3lnx}{\sqrt{x^2+1}}dx\)
12)\(\int\frac{xe^x}{_{ }\left(e^x+1\right)^2}dx\)
13) \(\int\frac{xln\left(x+\sqrt{1+x^2}\right)}{x+\sqrt{1+x^2}}dx\)
giúp mk đc con nào thì giúp nha
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 \(I=\frac{-1}{2\sqrt{3}}v+c\)
Câu 6)
\(I=-\int \frac{\left ( 1-\frac{1}{x^2} \right )dx}{x^2+2+\frac{1}{x^2}}=-\int \frac{d\left ( x+\frac{1}{x} \right )}{\left ( x+\frac{1}{x} \right )^2}=-\frac{1}{x+\frac{1}{x}}+c=-\frac{x}{x^2+1}+c\)
Câu 8)
\(I=\int \ln \left(\frac{x+1}{x-1}\right)dx=\int \ln (x+1)dx-\int \ln (x-1)dx\)
\(\Leftrightarrow I=\int \ln (x+1)d(x+1)-\int \ln (x-1)d(x-1)\)
Xét \(\int \ln tdt\) ta có:
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln tdt=t\ln t-\int dt=t\ln t-t+c\)
\(\Rightarrow I=(x+1)\ln (x+1)-(x+1)-(x-1)\ln (x-1)+x-1+c\)
\(\Leftrightarrow I=(x+1)\ln(x+1)-(x-1)\ln(x-1)+c\)
Tính các nguyên hàm.
a)\(\int\dfrac{2dx}{x^2-5x}=A\ln\left|x\right|+B\ln\left|x-5\right|+C\) . Tìm 2A-3B.
b)\(\int\dfrac{x^3-1}{x+1}\)dx=\(Ax^3-Bx^2+x+E\ln\left|x+1\right|+C\).Tính A-B+E
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) \(\int\dfrac{x^3-1}{x+1}dx=\int\dfrac{x^3+1-2}{x+1}dx=\int\left(x^2-x+1-\dfrac{2}{x+1}\right)dx=\dfrac{1}{3}x^3-\dfrac{1}{2}x^2+x-2ln\left|x+1\right|+C\)
\(\Rightarrow A=\dfrac{1}{3};B=\dfrac{1}{2};E=-2\Rightarrow A-B+E=-\dfrac{13}{6}\)
\(\int\dfrac{1}{cosx.cos\left(x+\dfrac{\pi}{4}\right)}dx\)
\(\int\dfrac{1}{x^3\left(1+x^2\right)}dx=\dfrac{a}{x^2}+blnx+cln\left(1+x^2\right).S=a+b+c=?\)
\(\int\dfrac{5-3x}{\left(x^2-5x+6\right)\left(x^2-2x+1\right)}dx=\dfrac{a}{x-1}-ln\left(\dfrac{x-b}{x-c}\right)+C.P=2a+b\)
Biến đổi: ʃ\(\int\dfrac{1dx}{cosx\dfrac{\sqrt{2}}{2}\left(cosx-sinx\right)}=\int\dfrac{\sqrt{2}dx}{cos^2x\left(1-tanx\right)}=\int\dfrac{\sqrt{2}d\left(tanx\right)}{1-tanx}=-\sqrt{2}\ln trituyetdoi\left(1-tanx\right)\)
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Áp dụng phương pháp tính tích phân, hãy tính các tích phân sau :
a) \(\int\limits^{\dfrac{\pi}{2}}_0x\cos2xdx\)
b) \(\int\limits^{\ln2}_0xe^{-2x}dx\)
c) \(\int\limits^1_0\ln\left(2x+1\right)dx\)
d) \(\int\limits^3_2\left|\ln\left(x-1\right)-\ln\left(x+1\right)\right|dx\)
e) \(\int\limits^2_{\dfrac{1}{2}}\left(1+x-\dfrac{1}{x}\right)e^{x+\dfrac{1}{x}}dx\)
g) \(\int\limits^{\dfrac{\pi}{2}}_0x\cos x\sin^2xdx\)
h) \(\int\limits^1_0\dfrac{xe^x}{\left(1+x\right)^2}dx\)
i) \(\int\limits^e_1\dfrac{1+x\ln x}{x}e^xdx\)
Sử dụng phương pháp tính nguyên hàm từng phần, hãy tính :
a) \(\int x\ln\left(1+x\right)dx\)
b) \(\int\left(x^2+2x-1\right)e^xdx\)
c) \(\int x\sin\left(2x+1\right)dx\)
d) \(\int\left(1-x\right)\cos xdx\)
a) Áp dụng phương pháp tìm nguyên hàm từng phần:
Đặt u= ln(1+x)
dv= xdx
=> ,
Ta có: ∫xln(1+x)dx =
=
b) Cách 1: Tìm nguyên hàm từng phần hai lần:
Đặt u= (x2+2x -1) và dv=exdx
Suy ra du = (2x+2)dx, v = ex
. Khi đó:
∫(x2+2x - 1)exdx = (x2+2x - 1)exdx - ∫(2x+2)exdx
Đặt : u=2x+2; dv=exdx
=> du = 2dx ;v=ex
Khi đó:∫(2x+2)exdx = (2x+2)ex - 2∫exdx = ex(2x+2) – 2ex+C
Vậy
∫(x2+2x+1)exdx = ex(x2-1) + C
Cách 2: HD: Ta tìm ∫(x2-1)exdx. Đặt u = x2-1 và dv=exdx.
Đáp số : ex(x2-1) + C
c) Đáp số:
HD: Đặt u=x ; dv = sin(2x+1)dx
d) Đáp số : (1-x)sinx - cosx +C.
HD: Đặt u = 1 - x ;dv = cosxdx
Tính nguyên hàm I = \(\int\left(x^2+2x\right)ln\left(3x+1\right)dx\)
1, \(\int\dfrac{x}{1-cos2x}dx\)
2, \(\int cos2x.e^{3x}dx\)
3, \(\int\left(2x+1\right)ln^2dx\)
4, \(\int\left(2x-1\right)cosxdx\)
5, \(\int\left(x^2+x+1\right)e^xdx\)
6, \(\int\left(2x+1\right)ln\left(x+2\right)dx\)
\(I=\int\dfrac{x}{1-cos2x}dx=\int\dfrac{x}{2sin^2x}dx\)
Đặt \(\left\{{}\begin{matrix}u=\dfrac{x}{2}\\dv=\dfrac{1}{sin^2x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{2}\\v=-cotx\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{-x.cotx}{2}+\dfrac{1}{2}\int cotxdx=\dfrac{-x.cotx}{2}+\dfrac{1}{2}\int\dfrac{cosx.dx}{sinx}\)
\(=\dfrac{-x.cotx}{2}+\dfrac{1}{2}\int\dfrac{d\left(sinx\right)}{sinx}=\dfrac{-x.cotx}{2}+\dfrac{1}{2}ln\left|sinx\right|+C\)
2/ Câu 2 bữa trước làm rồi, bạn coi lại nhé
3/ \(I=\int\left(2x+1\right)ln^2xdx\)
Đặt \(\left\{{}\begin{matrix}u=ln^2x\\dv=\left(2x+1\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{2lnx}{x}dx\\v=x^2+x\end{matrix}\right.\)
\(\Rightarrow I=\left(x^2+x\right)ln^2x-\int\left(2x+2\right)lnxdx=\left(x^2+x\right)ln^2x-I_1\)
\(I_1=\int\left(2x+2\right)lnx.dx\) \(\Rightarrow\left\{{}\begin{matrix}u=lnx\\dv=\left(2x+2\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=x^2+2x\end{matrix}\right.\)
\(\Rightarrow I_1=\left(x^2+2x\right)lnx-\int\left(x+2\right)dx=\left(x^2+2x\right)ln-\dfrac{x^2}{2}+2x+C\)
\(\Rightarrow I=\left(x^2+x\right)ln^2x-\left(x^2+2x\right)lnx+\dfrac{x^2}{2}-2x+C\)
4/ \(I=\int\left(2x-1\right)cosx.dx\) \(\Rightarrow\left\{{}\begin{matrix}u=2x-1\\dv=cosx.dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2dx\\v=sinx\end{matrix}\right.\)
\(\Rightarrow I=\left(2x-1\right)sinx-2\int sinx.dx=\left(2x-1\right)sinx+2cosx+C\)
5/ \(I=\int\left(x^2+x+1\right)e^xdx\) \(\Rightarrow\left\{{}\begin{matrix}u=x^2+x+1\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\left(2x+1\right)dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I=\left(x^2+x+1\right)e^x-\int\left(2x+1\right)e^xdx\)
\(I_1=\int\left(2x+1\right)e^xdx\) \(\Rightarrow\left\{{}\begin{matrix}u=2x+1\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2dx\\v=e^x\end{matrix}\right.\)
\(\Rightarrow I_1=\left(2x+1\right)e^x-2\int e^xdx=\left(2x+1\right)e^x-2e^x+C=\left(2x-1\right)e^x+C\)
\(\Rightarrow I=\left(x^2+x+1\right)e^x-\left(2x-1\right)e^x+C=\left(x^2-x+2\right)e^x+C\)
6/ \(I=\int\left(2x+1\right).ln\left(x+2\right)dx\)
\(\Rightarrow\left\{{}\begin{matrix}u=ln\left(x+2\right)\\dv=\left(2x+1\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x+2}\\v=x^2+x\end{matrix}\right.\)
\(\Rightarrow I=\left(x^2+x\right)ln\left(x+2\right)-\int\dfrac{x^2+x}{x+2}dx\)
\(=\left(x^2+x\right)ln\left(x+2\right)-\int\left(x-1+\dfrac{2}{x+2}\right)dx\)
\(I=\left(x^2+x\right)ln\left(x+2\right)-\dfrac{x^2}{2}+x-2ln\left|x+2\right|+C\)
Tính các tích phân sau :
a) \(\int\limits^{\dfrac{\pi}{4}}_0\cos2x.\cos^2xdx\)
b) \(\int\limits^1_{\dfrac{1}{2}}\dfrac{e^x}{e^{2x}-1}dx\)
c) \(\int\limits^1_0\dfrac{x+2}{x^2+2x+1}\ln\left(x+1\right)dx\)
d) \(\int\limits^{\dfrac{\pi}{4}}_0\dfrac{x\sin x+\left(x+1\right)\cos x}{x\sin x+\cos x}dx\)
a)
Ta có \(A=\int ^{\frac{\pi}{4}}_{0}\cos 2x\cos^2xdx=\frac{1}{4}\int ^{\frac{\pi}{4}}_{0}\cos 2x(\cos 2x+1)d(2x)\)
\(\Leftrightarrow A=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos x(\cos x+1)dx=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos xdx+\frac{1}{8}\int ^{\frac{\pi}{2}}_{0}(\cos 2x+1)dx\)
\(\Leftrightarrow A=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin x+\frac{1}{16}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin 2x+\frac{1}{8}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|x=\frac{1}{4}+\frac{\pi}{16}\)
b)
\(B=\int ^{1}_{\frac{1}{2}}\frac{e^x}{e^{2x}-1}dx=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\left ( \frac{1}{e^x-1}-\frac{1}{e^x+1} \right )d(e^x)\)
\(\Leftrightarrow B=\frac{1}{2}\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left | \frac{e^x-1}{e^x+1} \right |\approx 0.317\)
c)
Có \(C=\int ^{1}_{0}\frac{(x+2)\ln(x+1)}{(x+1)^2}d(x+1)\).
Đặt \(x+1=t\)
\(\Rightarrow C=\int ^{2}_{1}\frac{(t+1)\ln t}{t^2}dt=\int ^{2}_{1}\frac{\ln t}{t}dt+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
\(=\int ^{2}_{1}\ln td(\ln t)+\int ^{2}_{1}\frac{\ln t}{t^2}dt=\frac{\ln ^22}{2}+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)
\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)
d)
\(D=\int ^{\frac{\pi}{4}}_{0}\frac{x\sin x+(x+1)\cos x}{x\sin x+\cos x}dx=\int ^{\frac{\pi}{4}}_{0}dx+\int ^{\frac{\pi}{4}}_{0}\frac{x\cos x}{x\sin x+\cos x}dx\)
Ta có:
\(\int ^{\frac{\pi}{4}}_{0}dx=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|x=\frac{\pi}{4}\)
\(\int ^{\frac{\pi}{4}}_{0}\frac{x\cos xdx}{x\sin x+\cos x}=\int ^{\frac{\pi}{4}}_{0}\frac{d(x\sin x+\cos x)}{x\sin x+\cos x}=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\ln |x\sin x+\cos x|\)
\(=\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)
Suy ra \(D=\frac{\pi}{4}+\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)