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a/ \(I=\int\frac{9x^2}{\sqrt{1-x^3}}dx\)
Đặt \(u=\sqrt{1-x^3}\Rightarrow u^2=1-x^3\Rightarrow2u.du=-3x^2dx\)
\(\Rightarrow9x^2dx=-6udu\)
\(\Rightarrow I=\int\frac{-6u.du}{u}=-6\int du=-6u+C=-6\sqrt{1-x^3}+C\)
b/ Đặt \(u=1+\sqrt{x}\Rightarrow du=\frac{dx}{2\sqrt{x}}\Rightarrow2du=\frac{dx}{\sqrt{x}}\)
\(\Rightarrow I=\int\frac{2du}{u^3}=2\int u^{-3}du=-u^{-2}+C=-\frac{1}{u^2}+C=-\frac{1}{\left(1+\sqrt{x}\right)^2}+C\)
c/ Đặt \(u=\sqrt{2x+3}\Rightarrow u^2=2x\Rightarrow\left\{{}\begin{matrix}x=\frac{u^2}{2}\\dx=u.du\end{matrix}\right.\)
\(\Rightarrow I=\int\frac{u^2.u.du}{2u}=\frac{1}{2}\int u^2du=\frac{1}{6}u^3+C=\frac{1}{6}\sqrt{\left(2x+3\right)^3}+C\)
d/ Đặt \(u=\sqrt{1+e^x}\Rightarrow u^2-1=e^x\Rightarrow2u.du=e^xdx\)
\(\Rightarrow I=\int\frac{\left(u^2-1\right).2u.du}{u}=2\int\left(u^2-1\right)du=\frac{2}{3}u^3-2u+C\)
\(=\frac{2}{3}\sqrt{\left(1+e^x\right)^2}-2\sqrt{1+e^x}+C\)
e/ Đặt \(u=\sqrt[3]{1+lnx}\Rightarrow u^3=1+lnx\Rightarrow3u^2du=\frac{dx}{x}\)
\(\Rightarrow I=\int u.3u^2du=3\int u^3du=\frac{3}{4}u^4+C=\frac{3}{4}\sqrt[3]{\left(1+lnx\right)^4}+C\)
f/ \(I=\int cosx.sin^3xdx\)
Đặt \(u=sinx\Rightarrow du=cosxdx\)
\(\Rightarrow I=\int u^3du=\frac{1}{4}u^4+C=\frac{1}{4}sin^4x+C\)
1) Đặt \(t=1+\sqrt{x-1}\Leftrightarrow x=\left(t-1\right)^2+1\forall t\ge1\Rightarrow dx=d\left(t-1\right)^2=2dt\)
\(\Rightarrow I_1=\int\frac{\left(t-1\right)^2+1}{t}\cdot2dt=2\int\frac{t^2-2t+2}{t}dt=2\int\left(t-2+\frac{2}{t}\right)dt\\ =t^2-4t+4lnt+C\)
Thay x vào ta có...
2) \(I_2=\int\frac{2sinx\cdot cosx}{cos^3x-\left(1-cos^2x\right)-1}dx=\int\frac{-2cosx\cdot d\left(cosx\right)}{cos^3x+cos^2x-2}=\int\frac{-2t\cdot dt}{t^3+t-2}\)
\(I_2=\int\frac{-2t}{\left(t-1\right)\left(t^2+2t+2\right)}dt=-\frac{2}{5}\int\frac{dt}{t-1}+\frac{1}{5}\int\frac{2t+2}{t^2+2t+2}dt-\frac{6}{5}\int\frac{dt}{\left(t+1\right)^2+1}\)
Ta có:
\(\int\frac{2t+2}{t^2+2t+2}dt=\int\frac{d\left(t^2+2t+2\right)}{t^2+2t+2}=ln\left(t^2+2t+2\right)+C\)
\(\int\frac{dt}{\left(t+1\right)^2+1}=\int\frac{\frac{1}{cos^2m}}{tan^2m+1}dm=\int dm=m+C=arctan\left(t+1\right)+C\)
Thay x vào, ta có....
8.
\(I=\int sinx.cos2xdx=\int\left(2cos^2x-1\right)sinxdx\)
\(=\int\left(1-2cos^2x\right)d\left(cosx\right)=cosx-\frac{2}{3}cos^3x+C\)
9.
\(I=\int\frac{sin2x}{1+cos^2x}dx=-\int\frac{2\left(-sinx\right).cosx}{1+cos^2x}dx=-\int\frac{d\left(cos^2x\right)}{1+cos^2x}\)
\(=-ln\left|1+cos^2x\right|+C\)
6.
\(I=\int cos^3xdx=\int\left(1-sin^2x\right)cosxdx\)
\(=\int\left(1-sin^2x\right)d\left(sinx\right)=sinx-\frac{1}{3}sin^3x+C\)
7.
\(I=\int sin^2x.cos^3xdx=\int sin^2x\left(1-sin^2x\right)cosxdx\)
\(=\int\left(sin^2x-sin^4x\right)d\left(sinx\right)=\frac{1}{3}sin^3x-\frac{1}{5}sin^5x+C\)
\(I=\int\dfrac{x^3dx}{\left(x^8-4\right)^2}\)
Đặt \(x^4=t\Rightarrow x^3dx=\dfrac{1}{4}dt\Rightarrow I=\dfrac{1}{4}\int\dfrac{dt}{\left(t^2-2\right)^2}=\dfrac{1}{4}\int\dfrac{dt}{\left(t-\sqrt{2}\right)^2\left(t+\sqrt{2}\right)^2}\)
\(=\dfrac{1}{32}\int\left(\dfrac{1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}\right)^2dt=\dfrac{1}{32}\int\left(\dfrac{1}{\left(t-\sqrt{2}\right)^2}+\dfrac{1}{\left(t+\sqrt{2}\right)^2}-\dfrac{2}{\left(t+\sqrt{2}\right)\left(t-\sqrt{2}\right)}\right)dt\)
\(=\dfrac{1}{32}\int\left(\dfrac{1}{\left(t-\sqrt{2}\right)^2}+\dfrac{1}{\left(t+\sqrt{2}\right)^2}-\dfrac{1}{\sqrt{2}}\left(\dfrac{1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}\right)\right)dt\)
\(=\dfrac{1}{32}\left(\dfrac{-1}{t-\sqrt{2}}-\dfrac{1}{t+\sqrt{2}}-\dfrac{1}{\sqrt{2}}ln\left|\dfrac{t-\sqrt{2}}{t+\sqrt{2}}\right|\right)+C\)
\(=\dfrac{1}{32}\left(\dfrac{-1}{x^4-\sqrt{2}}-\dfrac{1}{x^4+\sqrt{2}}-\dfrac{1}{\sqrt{2}}ln\left|\dfrac{x^4-\sqrt{2}}{x^4+\sqrt{2}}\right|\right)+C\)
2/ \(I=\int\dfrac{\left(2x+1\right)dx}{\left(x^2+x-1\right)\left(x^2+x+3\right)}=\dfrac{1}{4}\int\left(\dfrac{1}{x^2+x-1}-\dfrac{1}{x^2+x+3}\right)\left(2x+1\right)dx\)
\(=\dfrac{1}{4}\int\left(\dfrac{2x+1}{x^2+x-1}-\dfrac{2x+1}{x^2+x+3}\right)dx\)
\(=\dfrac{1}{4}\left(\int\dfrac{d\left(x^2+x-1\right)}{x^2+x-1}-\int\dfrac{d\left(x^2+x+3\right)}{x^2+x+3}\right)\)
\(=\dfrac{1}{4}ln\left|\dfrac{x^2+x-1}{x^2+x+3}\right|+C\)
3/ Đặt \(\sqrt[3]{x}=t\Rightarrow x=t^3\Rightarrow dx=3t^2dt\)
\(\Rightarrow I=\int\dfrac{3t^2.sint.dt}{t^2}=3\int sint.dt=-3cost+C=-3cos\left(\sqrt[3]{x}\right)+C\)
4/ \(I=\int\dfrac{dx}{1+cos^2x}=\int\dfrac{\dfrac{1}{cos^2x}dx}{\dfrac{1}{cos^2x}+1}\)
Đặt \(t=tanx\Rightarrow\left\{{}\begin{matrix}dt=\dfrac{1}{cos^2x}dx\\\dfrac{1}{cos^2x}=1+tan^2x=1+t^2\end{matrix}\right.\)
\(\Rightarrow I=\int\dfrac{dt}{1+t^2+1}=\int\dfrac{dt}{t^2+2}=\dfrac{1}{2}\int\dfrac{dt}{\left(\dfrac{t}{\sqrt{2}}\right)^2+1}\)
\(=\dfrac{1}{2}.\sqrt{2}.arctan\left(\dfrac{t}{\sqrt{2}}\right)+C=\dfrac{1}{\sqrt{2}}arctan\left(\dfrac{tanx}{\sqrt{2}}\right)+C\)
5/ \(I=\int\dfrac{sinx+cosx}{4+2sinx.cosx-sin^2x-cos^2x}dx=\int\dfrac{sinx+cosx}{4-\left(sinx-cosx\right)^2}dx\)
Đặt \(sinx-cosx=t\Rightarrow\left(cosx+sinx\right)dx=dt\)
\(\Rightarrow I=\int\dfrac{dt}{4-t^2}=-\int\dfrac{dt}{\left(t-2\right)\left(t+2\right)}=\dfrac{1}{4}\int\left(\dfrac{1}{t+2}-\dfrac{1}{t-2}\right)dt\)
\(=\dfrac{1}{4}ln\left|\dfrac{t+2}{t-2}\right|+C=\dfrac{1}{4}ln\left|\dfrac{sinx-cosx+2}{sinx-cosx-2}\right|+C\)
Ơ bài 1 nhầm số 4 thành số 2 rồi, bạn sửa lại 1 chút nhé :D
Còn 1 cách làm khác nữa là lượng giác hóa
Đặt \(x^4=2sint\Rightarrow x^3dx=\dfrac{1}{2}cost.dt\)
\(\Rightarrow I=\dfrac{1}{2}\int\dfrac{cost.dt}{\left(4sin^2t-4\right)^2}=\dfrac{1}{32}\int\dfrac{cost.dt}{cos^4t}=\dfrac{1}{32}\int\dfrac{dt}{cos^3t}\)
Đặt \(\left\{{}\begin{matrix}u=\dfrac{1}{cost}\\dv=\dfrac{dt}{cos^2t}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{sint.dt}{cos^2t}\\v=tant\end{matrix}\right.\)
\(\Rightarrow32I=\dfrac{tant}{cost}-\int\dfrac{tant.sint.dt}{cos^2t}=\dfrac{sint}{cos^2t}-\int\dfrac{sin^2t.dt}{cos^3t}\)
\(=\dfrac{sint}{1-sin^2t}-\int\dfrac{1-cos^2t}{cos^3t}dt=\dfrac{sint}{1-sin^2t}-\int\dfrac{dt}{cos^3t}+\int\dfrac{1}{cosx}dx\)
Chú ý rằng \(\int\dfrac{dt}{cos^3t}=32I\)
\(\Rightarrow32I=\dfrac{sint}{1-sin^2t}-32I+\int\dfrac{cost.dt}{cos^2t}\)
\(\Rightarrow64I=\dfrac{sint}{1-sin^2t}-\int\dfrac{d\left(sint\right)}{sin^2t-1}=\dfrac{sint}{1-sin^2t}-\dfrac{1}{2}ln\left|\dfrac{sint-1}{sint+1}\right|+C\)
\(\Rightarrow I=\dfrac{1}{64}\left(\dfrac{2x^4}{4-x^8}-\dfrac{1}{2}ln\left|\dfrac{x^4-2}{x^4+2}\right|\right)+C\)
Câu 1)
Ta có \(I=\int ^{1}_{0}\frac{dx}{\sqrt{3+2x-x^2}}=\int ^{1}_{0}\frac{dx}{4-(x-1)^2}\).
Đặt \(x-1=2\cos t\Rightarrow \sqrt{4-(x-1)^2}=\sqrt{4-4\cos^2t}=2|\sin t|\)
Khi đó:
\(I=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{d(2\cos t+1)}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{2\sin tdt}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}dt=\left.\begin{matrix} \frac{2\pi}{3}\\ \frac{\pi}{2}\end{matrix}\right|t=\frac{\pi}{6}\)
Câu 3)
\(K=\int ^{3}_{2}\ln (x^3-3x+2)dx=\int ^{3}_{2}\ln [(x+2)(x-1)^2]dx\)
\(=\int ^{3}_{2}\ln (x+2)d(x+2)+2\int ^{3}_{2}\ln (x-1)d(x-1)\)
Xét \(\int \ln tdt\): Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln t dt=t\ln t-t\)
\(\Rightarrow K=\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x+2)[\ln (x+2)-1]+2\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x-1)[\ln (x-1)-1]\)
\(=5\ln 5-4\ln 4-1+4\ln 2-2=5\ln 5-4\ln 2-3\)
Bài 2)
\(J=\int ^{1}_{0}x\ln (2x+1)dx\). Đặt \(\left\{\begin{matrix} u=\ln (2x+1)\\ dv=xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2dx}{2x+1}\\ v=\frac{x^2}{2}\end{matrix}\right.\)
Khi đó:
\(J=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2\ln (2x+1)}{2}-\int ^{1}_{0}\frac{x^2}{2x+1}dx\)\(=\frac{\ln 3}{2}-\frac{1}{4}\int ^{1}_{0}(2x-1+\frac{1}{2x+1})dx\)
\(=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2-x}{4}-\frac{1}{8}\int ^{1}_{0}\frac{d(2x+1)}{2x+1}=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\ln (2x+1)}{8}\)
\(=\frac{\ln 3}{2}-\frac{\ln 3}{8}=\frac{3\ln 3}{8}\)
1) Đặt \(2+lnx=t\Leftrightarrow x=e^{t-2}\Rightarrow dx=e^{t-2}dt\)
\(I_1=\int\left(\frac{t-2}{t}\right)^2\cdot e^{t-2}\cdot dt=\int\left(1-\frac{4}{t}+\frac{4}{t^2}\right)e^{t-2}dt\\ =\int e^{t-2}dt-4\int\frac{e^{t-2}}{t}dt+4\int\frac{e^{t-2}}{t^2}dt\)
Có:
\(4\int\frac{e^{t-2}}{t^2}dt=-4\int e^{t-2}\cdot d\left(\frac{1}{t}\right)=-\frac{4\cdot e^{t-2}}{t}+4\int\frac{e^{t-2}}{t}dt\\ \Leftrightarrow4\int\frac{e^{t-2}}{t^2}dt-4\int\frac{e^{t-2}}{t^{ }}dt=-\frac{4\cdot e^{t-2}}{t}\)
Vậy \(I_1=\int e^{t-2}dt-\frac{4\cdot e^{t-2}}{t}=e^{t-2}-\frac{4e^{t-2}}{t}+C\)
3) Đặt \(t=\sqrt{1+\sqrt[3]{x^2}}\Rightarrow t^2-1=\sqrt[3]{x^2}\Leftrightarrow x^2=\left(t^2-1\right)^3\)
\(d\left(x^2\right)=d\left[\left(t^2-1\right)^3\right]\Leftrightarrow2x\cdot dx=6t\left(t^2-1\right)^2\cdot dt\)
\(I_3=\int\frac{3t\left(t^2-1\right)^2}{t}dt=3\int\left(t^4-2t^2+1\right)dt=...\)
\(I_1=\int cos\left(\frac{\pi x}{2}\right)dx-\int\frac{2}{6x+5}dx=\frac{2}{\pi}\int cos\left(\frac{\pi x}{2}\right)d\left(\frac{\pi x}{2}\right)-\frac{1}{3}\int\frac{d\left(6x+5\right)}{6x+5}\)
\(=\frac{2}{\pi}sin\left(\frac{\pi x}{2}\right)-\frac{1}{3}ln\left|6x+5\right|+C\)
\(I_2=-\frac{1}{2}\int\left(4-x^4\right)^{\frac{1}{2}}d\left(4-x^4\right)=-\frac{1}{2}.\frac{\left(4-x^4\right)^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{-\sqrt{\left(4-x^4\right)^3}}{3}+C\)
\(I_3=2\int e^{\frac{1}{2}\left(4+x^2\right)}d\left(\frac{1}{2}\left(4+x^2\right)\right)=2e^{\frac{1}{2}\left(4+x^2\right)}+C=2\sqrt{e^{4+x^2}}+C\)
\(I_4=-\frac{1}{2}\int\left(1-x^2\right)^{\frac{1}{3}}d\left(1-x^2\right)=-\frac{1}{2}.\frac{\left(1-x^2\right)^{\frac{4}{3}}}{\frac{4}{3}}+C=-\frac{3}{8}\sqrt[3]{\left(1-x^2\right)^4}+C\)
\(I_5=\int e^{sinx}d\left(sinx\right)=e^{sinx}+C\)
\(I_6=\int\frac{d\left(1+sinx\right)}{1+sinx}=ln\left(1+sinx\right)+C\)
\(I_7=\int\left(x+1\right)\sqrt{x-1}dx\)
Đặt \(\sqrt{x-1}=t\Rightarrow x=t^2+1\Rightarrow dx=2tdt\)
\(\Rightarrow I_7=\int\left(t^2+2\right).t.2t.dt=\int\left(2t^4+4t^2\right)dt=\frac{2}{5}t^5+\frac{4}{3}t^3+C\)
\(=\frac{2}{5}\sqrt{\left(1-x\right)^5}+\frac{4}{3}\sqrt{\left(1-x\right)^3}+C\)
\(I_8=\int\left(2x+1\right)^{20}dx\)
Đặt \(2x+1=t\Rightarrow2dx=dt\Rightarrow dx=\frac{1}{2}dt\)
\(\Rightarrow I_8=\frac{1}{2}\int t^{20}dt=\frac{1}{42}t^{21}+C=\frac{1}{42}\left(2x+1\right)^{21}+C\)
\(I_9=-3\int\left(1-x^3\right)^{-\frac{1}{2}}d\left(1-x^3\right)=-3.\frac{\left(1-x^3\right)^{\frac{1}{2}}}{\frac{1}{2}}+C=-6\sqrt{1-x^3}+C\)
\(I_{10}=\int\frac{x}{\sqrt{2x+3}}dx\)
Đặt \(\sqrt{2x+3}=t\Rightarrow x=\frac{1}{2}t^2-\frac{3}{2}\Rightarrow dx=t.dt\)
\(\Rightarrow I_{10}=\int\frac{\frac{1}{2}t^2-\frac{3}{2}}{t}.t.dt=\frac{1}{2}\int\left(t^2-3\right)dt=\frac{2}{3}t^3-\frac{3}{2}t+C\)
\(=\frac{2}{3}\sqrt{\left(2x+3\right)^3}-\frac{3}{2}\sqrt{2x+3}+C\)
a/ \(I=\int cos\left(\frac{\pi x}{2}\right)dx-\int\frac{2}{6x+5}dx=I_1+I_2\)
Xét \(I_1=\int cos\left(\frac{\pi x}{2}\right)dx\)
Đặt \(u=\frac{\pi x}{2}\Rightarrow x=\frac{2}{\pi}u\Rightarrow dx=\frac{2}{\pi}du\)
\(\Rightarrow I_1=\int cosu.\frac{2}{\pi}du=\frac{2}{\pi}\int cosu.du=\frac{2}{\pi}sinu+C=\frac{2}{\pi}sin\left(\frac{\pi x}{2}\right)+C\)
Xét \(I_2=\int\frac{2}{6x+5}dx\)
Đặt \(u=6x+5\Rightarrow x=\frac{1}{6}u-\frac{5}{6}\Rightarrow dx=\frac{1}{6}du\)
\(\Rightarrow I_2=\int\frac{2}{u}.\frac{1}{6}du=\frac{1}{3}\int\frac{du}{u}=\frac{1}{3}ln\left|u\right|+C=\frac{1}{3}ln\left|6x+5\right|+C\)
\(\Rightarrow I=I_1+I_2=\frac{2}{\pi}sin\left(\frac{\pi}{2}x\right)-\frac{1}{3}ln\left|6x+5\right|+C\)
b/ Đặt \(\sqrt{4-x^4}=u\Rightarrow x^4=4-u^2\Rightarrow4x^3dx=-2udu\)
\(\Rightarrow2x^3dx=-udu\)
\(\Rightarrow I=\int u.\left(-u.du\right)=-\int u^2du=-\frac{1}{3}u^3+C\)
\(=-\frac{1}{3}\sqrt{\left(4-x^4\right)^3}+C\)
c/ \(I=\int2e^{\frac{1}{2}\left(4+x^2\right)}.xdx\)
Đặt \(u=\frac{1}{2}\left(4+x^2\right)=2+\frac{x^2}{2}\Rightarrow xdx=du\)
\(\Rightarrow I=\int2e^udu=2e^u+C=2e^{\frac{1}{2}\left(4+x^2\right)}+C=2\sqrt{e^{4+x^4}}+C\)
d/ Đặt \(\sqrt[3]{1-x^2}=u\Rightarrow x^2=1-u^3\Rightarrow2xdx=-3u^2du\)
\(\Rightarrow xdx=-\frac{3}{2}u^2du\)
\(\Rightarrow I=\int u.\left(-\frac{3}{2}u^2du\right)=-\frac{3}{2}\int u^3du=-\frac{3}{8}u^4+C\)
\(=-\frac{3}{8}\sqrt[3]{\left(1-x^2\right)^4}+C\)
e/ Đặt \(sinx=u\Rightarrow cosxdx=du\)
\(\Rightarrow I=\int e^udu=e^u+C=e^{sinx}+C\)
f/ Đặt \(1+sinx=u\Rightarrow cosxdx=du\)
\(\Rightarrow I=\int\frac{du}{u}=ln\left|u\right|+C=ln\left|1+sinx\right|+C\)
Do \(1+sinx>0\) khi mẫu xác định nên có trị tuyệt đối hay ko cũng được
Câu 7;8;10 thay t bằng u là được
Câu 9:
Đặt \(\sqrt{1-x^3}=u\Rightarrow x^3=1-u^2\Rightarrow3x^2dx=-2udu\)
\(\Rightarrow9x^2dx=-6udu\)
\(\Rightarrow I=\int\frac{-6udu}{u}=-6\int du=-6u+C=-6\sqrt{1-x^3}+C\)