Evaluate the integrals. intfrac{tan^{-1}x}{x^2}dx

djeljenike 2020-11-01 Answered
Evaluate the integrals. tan1xx2dx
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Expert Answer

pierretteA
Answered 2020-11-02 Author has 102 answers
Given
tan1xx2dx
Apply integration by parts:fg=fgfg
f=tan1(x), g=1x2
f=1x2+1, g=1x
fg=tan1(x)x1x(x2+1)dx
1x(x2+1)dx=1x(x2+1)dx
The integral solce by expanding the fraction by1x3
1x(x2+1)dx=1(1x2+1)x3dx
Substitute u=1x2+1dudx=2x3dx=x32du
1(1x2+1)x3dx=1(u)x3×x32du=121udu
=ln(u)2
Undo substitution: u=1x2+1
=ln(1x2+1)2
1x(x2+1)dx=ln(1x2+1)2
tan1(x)x1x(x2+1)dx=tan1(x)xln(1x2+1)2+C
tan1xx2dx=tan1(x)xln(1x2+1)2+C
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