How do you find the integration of log⁡x

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Jack Maxson · Accepted Answer

Explanation:∫log⁡(x)dx=∫ln⁡(x)ln⁡(10)dx=1ln⁡(10)∫ln⁡(x)dx∫ln⁡(x)dxUsing the integration by parts :∫f(x)g′(x)dx=[f(x)g(x)]−∫f′(x)g(x)dxThere: f(x)=ln⁡(x),f′(x)=1x,g(x)=x,g′(x)=1So: ∫log⁡(x)dx=1ln⁡(10)(xln⁡(x)−∫dx)So: ∫log⁡(x)dx=1ln⁡(10)(xln⁡(x)−x)+C=xln⁡(10)(ln⁡(x)−1)+CIn general, ∫logn(x)dx=xln⁡(n)(ln⁡(x)−x)+C=xln⁡(10)(ln⁡(x)−1)+CIn general, ∫logn(x)dx=xln⁡(n)(ln⁡(x)−1)+C

Jillian Edgerton · Accepted Answer

Explanation:  Remember that:  loga(b)=logc(b)logc(a)  ⇒log10(x)=ln⁡(x)ln⁡(10)  We now have:  ∫ln⁡(x)⋅1ln⁡(10)dx  1ln⁡(10)∫ln⁡(x)dx  Integration by parts:  ∫udv=uv−∫vdu  We let:  u=ln⁡(x)  dv=1  ⇒du=ddx(ln⁡(x))  ⇒du=1x  ⇒v=∫1dx  ⇒v=x  ⇒1ln⁡(10)⋅[xln⁡(x)−∫x⋅1xdx]  ⇒1ln⁡(10)⋅[xln⁡(x)−∫1dx]  ⇒1ln⁡(10)⋅[xln⁡(x)−x]  ⇒xln⁡(x)−xln⁡(10)  ⇒xln⁡(x)−xln⁡(10)  ⇒x(ln⁡(x)−1)ln⁡(10) Do you C why this is incomplete?  ⇒∫log⁡(x)dx=x(ln⁡(x)−1)ln⁡(10)+C

How do you find the integration of \log x

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