1 \(\int x^3arcsin\frac{1}{x} dx\)
2 \(\int arctg\sqrt{x}dx\)
3 \(\int(arccosx)^2 dx\)
4 \(\int xarctgx^2 dx\)
Całki
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2
\(\int arctg\sqrt{x}dx= \left( \sqrt{x}=t\\x=t^2\\ \frac{dx}{dt}=2t\\dx=2tdt \right)= \int 2t \cdot arctg t dt=\int \left(t^2 \right)' \cdot arctg t dt=
t^2 arctg t -\int \frac{t^2}{1+t^2} dt=
t^2 arctg t -\int \frac{1+t^2-1}{1+t^2} dt=
t^2 arctg t -\int 1-\frac{1}{1+t^2} dt=
t^2 arctg t -t+arctgt+C=
x arctg \sqrt{x} - \sqrt{x} +arctg \sqrt{x} +C\)
\(\int arctg\sqrt{x}dx= \left( \sqrt{x}=t\\x=t^2\\ \frac{dx}{dt}=2t\\dx=2tdt \right)= \int 2t \cdot arctg t dt=\int \left(t^2 \right)' \cdot arctg t dt=
t^2 arctg t -\int \frac{t^2}{1+t^2} dt=
t^2 arctg t -\int \frac{1+t^2-1}{1+t^2} dt=
t^2 arctg t -\int 1-\frac{1}{1+t^2} dt=
t^2 arctg t -t+arctgt+C=
x arctg \sqrt{x} - \sqrt{x} +arctg \sqrt{x} +C\)
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\(\int x^3 arcsin\frac{1}{x}dx=\int \frac{1}{4}\(x^4\)^{'} arcsin\frac{1}{x}dx=
=\frac{1}{4}x^4arcsin\frac{1}{x}-\frac{1}{4}\int x^4 \frac{1}{\sqrt{1-\frac{1}{x^2}}} \(-\frac{1}{x^2}\)dx=
=\frac{1}{4}x^4arcsin\frac{1}{x}+\frac{1}{4}\int x^2 \frac{x}{\sqrt{x^2-1}} dx=
=\frac{1}{4}x^4arcsin\frac{1}{x}+\frac{1}{8}\int x^2 \frac{2x}{\sqrt{x^2-1}} dx \Rightarrow \[x^2-1=t\\ \frac{dt}{dx} =2x\\ x^2=t+1\] \Rightarrow
=\frac{1}{4}x^4arcsin\frac{1}{x}+\frac{1}{8}\int (t+1) \frac{1}{\sqrt{t}} dt=
=\frac{1}{4}x^4arcsin\frac{1}{x}+\frac{1}{8}\int t^{\frac{1}{2}}+t^{-\frac{1}{2}} dt=
=\frac{1}{4}x^4arcsin\frac{1}{x}+\frac{1}{8} \cdot \frac{2}{3}t^{\frac{3}{2}}+\frac{1}{8}\cdot 2t^{\frac{1}{2}} +C=
=\frac{1}{4}x^4arcsin\frac{1}{x}+\frac{1}{12}\(x^2-1\)^{\frac{3}{2}}+\frac{1}{4}\(x^2-1\)^{\frac{1}{2}}+C\)
=\frac{1}{4}x^4arcsin\frac{1}{x}-\frac{1}{4}\int x^4 \frac{1}{\sqrt{1-\frac{1}{x^2}}} \(-\frac{1}{x^2}\)dx=
=\frac{1}{4}x^4arcsin\frac{1}{x}+\frac{1}{4}\int x^2 \frac{x}{\sqrt{x^2-1}} dx=
=\frac{1}{4}x^4arcsin\frac{1}{x}+\frac{1}{8}\int x^2 \frac{2x}{\sqrt{x^2-1}} dx \Rightarrow \[x^2-1=t\\ \frac{dt}{dx} =2x\\ x^2=t+1\] \Rightarrow
=\frac{1}{4}x^4arcsin\frac{1}{x}+\frac{1}{8}\int (t+1) \frac{1}{\sqrt{t}} dt=
=\frac{1}{4}x^4arcsin\frac{1}{x}+\frac{1}{8}\int t^{\frac{1}{2}}+t^{-\frac{1}{2}} dt=
=\frac{1}{4}x^4arcsin\frac{1}{x}+\frac{1}{8} \cdot \frac{2}{3}t^{\frac{3}{2}}+\frac{1}{8}\cdot 2t^{\frac{1}{2}} +C=
=\frac{1}{4}x^4arcsin\frac{1}{x}+\frac{1}{12}\(x^2-1\)^{\frac{3}{2}}+\frac{1}{4}\(x^2-1\)^{\frac{1}{2}}+C\)
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\(\int x arctg x^2dx= \frac{1}{2} \int \(x^2\)^{'} arctg x^2dx =
=\frac{1}{2}x^2arctg x^2-\frac{1}{2} \int x^2 \frac{1}{1+x^4}2xdx =
=\frac{1}{2}x^2arctg x^2-\frac{1}{4} \int \frac{4x^3}{1+x^4}dx \Rightarrow \[t=x^4+1\\ \frac{dt}{dx}= 4x^3\] \Rightarrow
\frac{1}{2}x^2arctg x^2-\frac{1}{4} \int \frac{1}{t}dt =
= \frac{1}{2}x^2arctg x^2-\frac{1}{4} \ln \|t\|+C=
=\frac{1}{2}x^2arctg x^2-\frac{1}{4} \ln \|x^4+1\|+C\)
=\frac{1}{2}x^2arctg x^2-\frac{1}{2} \int x^2 \frac{1}{1+x^4}2xdx =
=\frac{1}{2}x^2arctg x^2-\frac{1}{4} \int \frac{4x^3}{1+x^4}dx \Rightarrow \[t=x^4+1\\ \frac{dt}{dx}= 4x^3\] \Rightarrow
\frac{1}{2}x^2arctg x^2-\frac{1}{4} \int \frac{1}{t}dt =
= \frac{1}{2}x^2arctg x^2-\frac{1}{4} \ln \|t\|+C=
=\frac{1}{2}x^2arctg x^2-\frac{1}{4} \ln \|x^4+1\|+C\)