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NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 11 2019
\(I_1=\int cos\left(\frac{\pi x}{2}\right)dx-\int\frac{2}{6x+5}dx=\frac{2}{\pi}\int cos\left(\frac{\pi x}{2}\right)d\left(\frac{\pi x}{2}\right)-\frac{1}{3}\int\frac{d\left(6x+5\right)}{6x+5}\)
\(=\frac{2}{\pi}sin\left(\frac{\pi x}{2}\right)-\frac{1}{3}ln\left|6x+5\right|+C\)
\(I_2=-\frac{1}{2}\int\left(4-x^4\right)^{\frac{1}{2}}d\left(4-x^4\right)=-\frac{1}{2}.\frac{\left(4-x^4\right)^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{-\sqrt{\left(4-x^4\right)^3}}{3}+C\)
\(I_3=2\int e^{\frac{1}{2}\left(4+x^2\right)}d\left(\frac{1}{2}\left(4+x^2\right)\right)=2e^{\frac{1}{2}\left(4+x^2\right)}+C=2\sqrt{e^{4+x^2}}+C\)
\(I_4=-\frac{1}{2}\int\left(1-x^2\right)^{\frac{1}{3}}d\left(1-x^2\right)=-\frac{1}{2}.\frac{\left(1-x^2\right)^{\frac{4}{3}}}{\frac{4}{3}}+C=-\frac{3}{8}\sqrt[3]{\left(1-x^2\right)^4}+C\)
\(I_5=\int e^{sinx}d\left(sinx\right)=e^{sinx}+C\)
\(I_6=\int\frac{d\left(1+sinx\right)}{1+sinx}=ln\left(1+sinx\right)+C\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 11 2019
\(I_7=\int\left(x+1\right)\sqrt{x-1}dx\)
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 \(\sqrt{2x+3}=t\Rightarrow x=\frac{1}{2}t^2-\frac{3}{2}\Rightarrow dx=t.dt\)
\(\Rightarrow I_{10}=\int\frac{\frac{1}{2}t^2-\frac{3}{2}}{t}.t.dt=\frac{1}{2}\int\left(t^2-3\right)dt=\frac{2}{3}t^3-\frac{3}{2}t+C\)
\(=\frac{2}{3}\sqrt{\left(2x+3\right)^3}-\frac{3}{2}\sqrt{2x+3}+C\)
HT
20 tháng 3 2021
Cách này hơi dài chút, nhưng nếu nghĩ ra cách hay hơn mình sẽ đề xuất nhe!
\(=\int\sin^5x.\left(2\sin x\cos x\right)^3.2xdx=16\int x.\sin^8x\cos^3xdx\)
\(\left\{{}\begin{matrix}u=x\\dv=\sin^8x.\cos^3xdx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\int\sin^8x.\cos^3xdx\end{matrix}\right.\)
\(I_1=\int\sin^8x\cos^3xdx=\int\sin^8x.\cos^2x.\cos xdx=\int\sin^8x.\left(1-\sin^2x\right)\cos xdx\)
\(t=\sin x\Rightarrow dt=\cos xdx\Rightarrow\int\sin^8x\left(1-\sin^2x\right)\cos xdx=\int(t^8-t^{10})dt=\dfrac{1}{9}t^9-\dfrac{1}{11}t^{11}=\dfrac{1}{9}\sin^9x-\dfrac{1}{11}\sin^{11}x\)
\(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\dfrac{1}{9}\sin^9x-\dfrac{1}{11}\sin^{11}x\end{matrix}\right.\)
\(\Rightarrow\dfrac{I}{16}=x.\left(\dfrac{1}{9}\sin^9x-11\sin^{11}x\right)-\int\left(\dfrac{1}{9}\sin^9x-\dfrac{1}{11}\sin^{11}x\right)dx\)
\(I_2=\int\left(\dfrac{1}{9}\sin^9x-\dfrac{1}{11}\sin^{11}x\right)dx=\dfrac{1}{9}\int\sin^9xdx-\dfrac{1}{11}\int\sin^{11}xdx\)
À thế này là xong rồi còn gì :) Bạn tự làm nốt nhé
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 3 2023
\(\int\dfrac{\left(1+lnx\right)^2}{x}dx=\int\left(1+lnx\right)^2d\left(1+lnx\right)=\dfrac{1}{3}\left(1+lnx\right)^3+C\)
Lời giải:
Ta có:
\(P=\int \frac{2xdx}{(x+1)(x^2+1)^2}=\int \frac{2x(x-1)dx}{(x^2-1)(x^2+1)^2}\)
\(=\int \frac{x(x-1)}{x^2+1}\left(\frac{1}{x^2-1}-\frac{1}{x^2+1}\right)dx\)
\(=\int \frac{x(x-1)}{(x^2+1)(x^2-1)}dx-\int \frac{x(x-1)}{(x^2+1)^2}dx=M-N\)
Xét M
\(M=\int \frac{x(x-1)}{(x^2+1)(x^2-1)}dx=\int \frac{x(x-1)}{2}\left(\frac{1}{x^2-1}-\frac{1}{x^2+1}\right)dx\)
\(=\int \frac{x}{2(x+1)}dx-\int \frac{x(x-1)}{2(x^2+1)}dx\)
\(=\frac{1}{2}\int (1-\frac{1}{x+1})dx-\frac{1}{2}\int (1-\frac{x+1}{x^2+1})dx\)
\(=\frac{1}{2}\int dx-\frac{1}{2}\int \frac{d(x+1)}{x+1}-\frac{1}{2}\int dx+\frac{1}{2}\int \frac{(x+1)dx}{x^2+1}\)
\(=-\frac{1}{2}\ln |x+1|+\frac{1}{2}\int \frac{(x+1)dx}{x^2+1}\)
Xét N
Đặt \(\left\{\begin{matrix} u=x-1\\ dv=\frac{xdx}{(x^2+1)^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=dx\\ v=\int \frac{xdx}{(x^2+1)^2}=\frac{1}{2}\int \frac{d(x^2+1)}{(x^2+1)^2}=\frac{-1}{2(x^2+1)}\end{matrix}\right.\)
\(\Rightarrow N=\frac{1-x}{2(x^2+1)}+\int \frac{1}{2(x^2+1)}dx\)
Do đó: \(P=M-N=-\frac{1}{2}\ln |x+1|+\frac{x-1}{2(x^2+1)}+\frac{1}{2}\int \frac{xdx}{x^2+1}\)
\(=\frac{-1}{2}\ln |x+1|+\frac{x-1}{2(x^2+1)}+\frac{1}{4}\int \frac{d(x^2+1)}{x^2+1}\)
\(=\frac{-1}{2}\ln |x+1|+\frac{x-1}{2(x^2+1)}+\frac{1}{4}\ln |x^2+1|+c\)