Evaluate the integrals. ∫tan−1⁡xx2dx

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pierretteA · Accepted Answer

Given∫tan−1⁡xx2dxApply integration by parts:∫f⋅g′=fg−∫f′⋅gf=tan−1⁡(x), g′=1x2f′=1x2+1, g=−1x∫f⋅g′=−tan−1⁡(x)x−∫−1x(x2+1)dx∫−1x(x2+1)dx=−∫1x(x2+1)dxThe integral solce by expanding the fraction by1x3∫1x(x2+1)dx=∫1(1x2+1)x3dxSubstitute u=1x2+1→dudx=−2x3→dx=−x32du∫1(1x2+1)x3dx=∫1(u)x3×−x32du=−12∫1udu=−ln⁡(u)2Undo substitution: u=1x2+1=−ln⁡(1x2+1)2⇒−∫1x(x2+1)dx=ln⁡(1x2+1)2⇒−tan−1⁡(x)x−∫−1x(x2+1)dx=tan−1⁡(x)x−ln⁡(1x2+1)2+C∫tan−1⁡xx2dx=tan−1⁡(x)x−ln⁡(1x2+1)2+C

Evaluate the integrals. ∫tan−1⁡xx2dx

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