The formula for partial integration. If u = F(x) and v =
G(x) and they are differentiable functions than:
$$\int{udv=uv-\int{vdu}}$$
1.$$\int{\ln xdx=\left| \begin{matrix}
u=\ln x/' & dv=dx/\int{{}} \\
du=\frac{dx}{x} & v=x \\
\end{matrix} \right|}=uv-\int{vdu=x\ln x-\int{x\frac{dx}{x}}}=x\ln x-\int{dx}=x\ln x-x+C$$ 2.$$\int{\arctan xdx=\left| \begin{matrix}
u=\arctan x/' & dv=dx/\int{{}} \\
du=\frac{dx}{1+{{x}^{2}}} & v=x \\
\end{matrix} \right|}=uv-\int{vdu}=x\arctan x-\int{\frac{xdx}{1+{{x}^{2}}}}=$$
$$=x\arctan x-\left| \begin{align}
& t=1+{{x}^{2}}/' \\
& dt=2xdx \\
& xdx=\frac{dt}{2} \\
\end{align} \right|=x\arctan x-\frac{1}{2}\int{\frac{dt}{t}}=x\arctan x-\frac{1}{2}\ln \left| t \right|+C=x\arctan x-\frac{1}{2}\ln \left| 1+{{x}^{2}} \right|+C$$
3.$$\int{\arcsin xdx=\left| \begin{matrix} u=\arcsin x/' & dv=dx/\int{{}} \\ du=\frac{dx}{\sqrt{1-{{x}^{2}}}} & v=x \\ \end{matrix} \right|}=uv-\int{vdu}=x\arcsin x-\int{\frac{xdx}{\sqrt{1-{{x}^{2}}}}}=$$ $$=x\arcsin x-\left| \begin{align} & t=1-{{x}^{2}}/' \\ & dt=-2xdx \\ & -\frac{dt}{2}=xdx \\ \end{align} \right|=x\arcsin x+\frac{1}{2}\int{\frac{dt}{\sqrt{t}}}=x\arcsin x+\frac{1}{2}\frac{{{t}^{-\frac{1}{2}+1}}}{-\frac{1}{2}+1}+C=x\arcsin x+\frac{1}{2}2\sqrt{t}+C$$ $$=x\arcsin x+\sqrt{1-{{x}^{2}}}+C$$
4.$$\begin{align} & \int{x\sin xdx}=\left| \begin{matrix} u=x/' & dv=\sin xdx/\int{{}} \\ du=dx & v=-\cos x \\ \end{matrix} \right|=uv-\int{vdu}=-x\cos x+\int{\cos x}dx= \\ & =-x\cos x+\sin x+C \\ \end{align}$$
5.$$\int{x\cos 3x}dx=\left| \begin{matrix} u=x/' & dv=\cos 3xdx/\int{{}} \\ du=dx & v=\int{\cos 3xdx} \\ \end{matrix} \right|=$$ $$v=\int{\cos 3xdx}=\left| \begin{align} & t=3x/' \\ & dt=3dx \\ & dx=\frac{1}{3}dt \\ \end{align} \right|=\frac{1}{3}\int{\cos t}dt=\frac{1}{3}\sin 3x$$ $$\int{x\cos 3x}dx=uv-\int{vdu}=\frac{x}{3}\sin 3x-\frac{1}{3}\int{\sin 3x}dx=\frac{x\sin 3x}{3}+\frac{\cos 3x}{9}+C$$
6.$$\begin{align} & \int{\frac{x}{{{e}^{-x}}}dx=\int{x{{e}^{x}}dx}=\left| \begin{matrix} u=x/' & dv={{e}^{x}}dx/\int{{}} \\ du=dx & v={{e}^{x}} \\ \end{matrix} \right|}=uv-\int{vdu}= \\ & =x{{e}^{x}}-\int{{{e}^{x}}}dx=x{{e}^{x}}-{{e}^{x}}+C \\ \end{align}$$ $$\int{x{{2}^{-x}}dx=\left| \begin{matrix} u=x/' & dv={{2}^{-x}}dx/\int{{}} \\ du=dx & v=\int{{{2}^{-x}}dx} \\ \end{matrix} \right|}$$ $$v=\left| \begin{align} & t=-x/' \\ & dt=-dx \\ & dx=-dt \\ \end{align} \right|=-\int{{{2}^{t}}dt}=-\frac{{{2}^{-x}}}{\ln 2}$$ $$\int{x{{2}^{-x}}dx=}uv-\int{vdu}=-\frac{x{{2}^{-x}}}{\ln 2}+\int{\frac{{{2}^{-x}}}{\ln 2}}dx=-\frac{x{{2}^{-x}}}{\ln 2}-\frac{{{2}^{-x}}}{{{\ln }^{2}}2}+C$$ $$\int{{{x}^{2}}{{e}^{3x}}dx}=\left| \begin{matrix} u={{x}^{2}}/' & dv={{e}^{3x}}dx/\int{{}} \\ du=2xdx & v=\int{{{e}^{3x}}dx} \\ \end{matrix} \right|=$$ $$v=\int{{{e}^{3x}}dx}=\left| \begin{align} & t=3x/' \\ & dt=3dx/:3 \\ & \frac{1}{3}dt=dx \\ \end{align} \right|=\frac{1}{3}\int{{{e}^{t}}dt}=\frac{1}{3}{{e}^{3x}}$$ $$\int{{{x}^{2}}{{e}^{3x}}dx}=uv-\int{vdu}=\frac{{{x}^{2}}}{3}{{e}^{3x}}-\frac{2}{3}\int{x{{e}^{3x}}dx}=\frac{{{x}^{2}}{{e}^{3x}}}{3}-\frac{2}{3}\left| \begin{matrix} u=x/' & dv={{e}^{3x}}dx/\int{{}} \\ du=dx & v=\int{{{e}^{3x}}dx} \\ \end{matrix} \right|=$$ $$v=\int{{{e}^{3x}}dx}=\left| \begin{align} & t=3x/' \\ & dt=3dx/:3 \\ & \frac{1}{3}dt=dx \\ \end{align} \right|=\frac{1}{3}\int{{{e}^{t}}dt}=\frac{1}{3}{{e}^{3x}}$$ $$\int{{{x}^{2}}{{e}^{3x}}dx}=\frac{{{x}^{2}}}{3}{{e}^{3x}}-\frac{2}{3}\left( uv-\int{vdu} \right)=\frac{{{x}^{2}}{{e}^{3x}}}{3}-\frac{2}{3}\left( \frac{{{x}^{2}}{{e}^{3x}}}{3}-\frac{1}{3}\int{{{e}^{3x}}dx} \right)=$$ $$\int{{{x}^{2}}{{e}^{3x}}dx}=\frac{{{x}^{2}}{{e}^{3x}}}{3}-\frac{2{{x}^{2}}{{e}^{3x}}}{9}+\frac{2}{9}\left| \begin{align} & t=3x/' \\ & dt=3dx/:3 \\ & \frac{1}{3}dt=dx \\ \end{align} \right|=$$ $$\int{{{x}^{2}}{{e}^{3x}}dx}=\frac{{{x}^{2}}{{e}^{3x}}}{3}-\frac{2{{x}^{2}}{{e}^{3x}}}{9}+\frac{2}{9}\frac{1}{3}{{e}^{3x}}+C=\frac{{{x}^{2}}{{e}^{3x}}}{3}-\frac{2{{x}^{2}}{{e}^{3x}}}{9}+\frac{2}{27}{{e}^{3x}}+C$$ $$\int{\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}dx}=\left| \begin{matrix} u={{x}^{2}}-2x+5/' & dv={{e}^{-x}}dx/\int{{}} \\ du=(2x-2)dx & v=\int{{{e}^{-x}}dx} \\ \end{matrix} \right|=$$ $$v=\int{{{e}^{-x}}dx}=\left| \begin{align} & t=-x/' \\ & dt=-dx \\ & -dt=dx \\ \end{align} \right|=-{{e}^{-x}}$$ $$\int{\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}dx}=uv-\int{vdu}=-\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}+\int{(2x-2)}{{e}^{-x}}dx=$$ $$\int{\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}dx}=-\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}+\left| \begin{matrix} u=2x-2/' & dv={{e}^{-x}}dx/\int{{}} \\ du=2dx & v=\int{{{e}^{-x}}dx} \\ \end{matrix} \right|=$$ $$v=\int{{{e}^{-x}}dx}=\left| \begin{align} & t=-x/' \\ & dt=-dx \\ & -dt=dx \\ \end{align} \right|=-{{e}^{-x}}$$ $$\int{\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}dx}=-\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}-\left( 2x-2 \right){{e}^{-x}}+2\int{{{e}^{-x}}}dx=-\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}-\left( 2x-2 \right){{e}^{-x}}-2{{e}^{-x}}+C$$
3.$$\int{\arcsin xdx=\left| \begin{matrix} u=\arcsin x/' & dv=dx/\int{{}} \\ du=\frac{dx}{\sqrt{1-{{x}^{2}}}} & v=x \\ \end{matrix} \right|}=uv-\int{vdu}=x\arcsin x-\int{\frac{xdx}{\sqrt{1-{{x}^{2}}}}}=$$ $$=x\arcsin x-\left| \begin{align} & t=1-{{x}^{2}}/' \\ & dt=-2xdx \\ & -\frac{dt}{2}=xdx \\ \end{align} \right|=x\arcsin x+\frac{1}{2}\int{\frac{dt}{\sqrt{t}}}=x\arcsin x+\frac{1}{2}\frac{{{t}^{-\frac{1}{2}+1}}}{-\frac{1}{2}+1}+C=x\arcsin x+\frac{1}{2}2\sqrt{t}+C$$ $$=x\arcsin x+\sqrt{1-{{x}^{2}}}+C$$
4.$$\begin{align} & \int{x\sin xdx}=\left| \begin{matrix} u=x/' & dv=\sin xdx/\int{{}} \\ du=dx & v=-\cos x \\ \end{matrix} \right|=uv-\int{vdu}=-x\cos x+\int{\cos x}dx= \\ & =-x\cos x+\sin x+C \\ \end{align}$$
5.$$\int{x\cos 3x}dx=\left| \begin{matrix} u=x/' & dv=\cos 3xdx/\int{{}} \\ du=dx & v=\int{\cos 3xdx} \\ \end{matrix} \right|=$$ $$v=\int{\cos 3xdx}=\left| \begin{align} & t=3x/' \\ & dt=3dx \\ & dx=\frac{1}{3}dt \\ \end{align} \right|=\frac{1}{3}\int{\cos t}dt=\frac{1}{3}\sin 3x$$ $$\int{x\cos 3x}dx=uv-\int{vdu}=\frac{x}{3}\sin 3x-\frac{1}{3}\int{\sin 3x}dx=\frac{x\sin 3x}{3}+\frac{\cos 3x}{9}+C$$
6.$$\begin{align} & \int{\frac{x}{{{e}^{-x}}}dx=\int{x{{e}^{x}}dx}=\left| \begin{matrix} u=x/' & dv={{e}^{x}}dx/\int{{}} \\ du=dx & v={{e}^{x}} \\ \end{matrix} \right|}=uv-\int{vdu}= \\ & =x{{e}^{x}}-\int{{{e}^{x}}}dx=x{{e}^{x}}-{{e}^{x}}+C \\ \end{align}$$ $$\int{x{{2}^{-x}}dx=\left| \begin{matrix} u=x/' & dv={{2}^{-x}}dx/\int{{}} \\ du=dx & v=\int{{{2}^{-x}}dx} \\ \end{matrix} \right|}$$ $$v=\left| \begin{align} & t=-x/' \\ & dt=-dx \\ & dx=-dt \\ \end{align} \right|=-\int{{{2}^{t}}dt}=-\frac{{{2}^{-x}}}{\ln 2}$$ $$\int{x{{2}^{-x}}dx=}uv-\int{vdu}=-\frac{x{{2}^{-x}}}{\ln 2}+\int{\frac{{{2}^{-x}}}{\ln 2}}dx=-\frac{x{{2}^{-x}}}{\ln 2}-\frac{{{2}^{-x}}}{{{\ln }^{2}}2}+C$$ $$\int{{{x}^{2}}{{e}^{3x}}dx}=\left| \begin{matrix} u={{x}^{2}}/' & dv={{e}^{3x}}dx/\int{{}} \\ du=2xdx & v=\int{{{e}^{3x}}dx} \\ \end{matrix} \right|=$$ $$v=\int{{{e}^{3x}}dx}=\left| \begin{align} & t=3x/' \\ & dt=3dx/:3 \\ & \frac{1}{3}dt=dx \\ \end{align} \right|=\frac{1}{3}\int{{{e}^{t}}dt}=\frac{1}{3}{{e}^{3x}}$$ $$\int{{{x}^{2}}{{e}^{3x}}dx}=uv-\int{vdu}=\frac{{{x}^{2}}}{3}{{e}^{3x}}-\frac{2}{3}\int{x{{e}^{3x}}dx}=\frac{{{x}^{2}}{{e}^{3x}}}{3}-\frac{2}{3}\left| \begin{matrix} u=x/' & dv={{e}^{3x}}dx/\int{{}} \\ du=dx & v=\int{{{e}^{3x}}dx} \\ \end{matrix} \right|=$$ $$v=\int{{{e}^{3x}}dx}=\left| \begin{align} & t=3x/' \\ & dt=3dx/:3 \\ & \frac{1}{3}dt=dx \\ \end{align} \right|=\frac{1}{3}\int{{{e}^{t}}dt}=\frac{1}{3}{{e}^{3x}}$$ $$\int{{{x}^{2}}{{e}^{3x}}dx}=\frac{{{x}^{2}}}{3}{{e}^{3x}}-\frac{2}{3}\left( uv-\int{vdu} \right)=\frac{{{x}^{2}}{{e}^{3x}}}{3}-\frac{2}{3}\left( \frac{{{x}^{2}}{{e}^{3x}}}{3}-\frac{1}{3}\int{{{e}^{3x}}dx} \right)=$$ $$\int{{{x}^{2}}{{e}^{3x}}dx}=\frac{{{x}^{2}}{{e}^{3x}}}{3}-\frac{2{{x}^{2}}{{e}^{3x}}}{9}+\frac{2}{9}\left| \begin{align} & t=3x/' \\ & dt=3dx/:3 \\ & \frac{1}{3}dt=dx \\ \end{align} \right|=$$ $$\int{{{x}^{2}}{{e}^{3x}}dx}=\frac{{{x}^{2}}{{e}^{3x}}}{3}-\frac{2{{x}^{2}}{{e}^{3x}}}{9}+\frac{2}{9}\frac{1}{3}{{e}^{3x}}+C=\frac{{{x}^{2}}{{e}^{3x}}}{3}-\frac{2{{x}^{2}}{{e}^{3x}}}{9}+\frac{2}{27}{{e}^{3x}}+C$$ $$\int{\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}dx}=\left| \begin{matrix} u={{x}^{2}}-2x+5/' & dv={{e}^{-x}}dx/\int{{}} \\ du=(2x-2)dx & v=\int{{{e}^{-x}}dx} \\ \end{matrix} \right|=$$ $$v=\int{{{e}^{-x}}dx}=\left| \begin{align} & t=-x/' \\ & dt=-dx \\ & -dt=dx \\ \end{align} \right|=-{{e}^{-x}}$$ $$\int{\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}dx}=uv-\int{vdu}=-\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}+\int{(2x-2)}{{e}^{-x}}dx=$$ $$\int{\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}dx}=-\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}+\left| \begin{matrix} u=2x-2/' & dv={{e}^{-x}}dx/\int{{}} \\ du=2dx & v=\int{{{e}^{-x}}dx} \\ \end{matrix} \right|=$$ $$v=\int{{{e}^{-x}}dx}=\left| \begin{align} & t=-x/' \\ & dt=-dx \\ & -dt=dx \\ \end{align} \right|=-{{e}^{-x}}$$ $$\int{\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}dx}=-\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}-\left( 2x-2 \right){{e}^{-x}}+2\int{{{e}^{-x}}}dx=-\left( {{x}^{2}}-2x+5 \right){{e}^{-x}}-\left( 2x-2 \right){{e}^{-x}}-2{{e}^{-x}}+C$$
No comments:
Post a Comment