By using integration by parts, evaluate the integral `int_0^1 x tan^-1 x dx.`

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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`int_0^1 x tan^-1 x dx `


`u = x^2/2`

`du = xdx`

`v = tan^(-1)x`

`dv = 1/(1+x^2)dx`

Using integral by parts,

`intvdu = uv-intudv`


`= [tan^(-1)x(x^2/2)]_0^1-int_0^1(x^2/2)/(1+x^2)dx`

`= [tan^(-1)x(x^2/2)]_0^1-1/2int_0^1(1+x^2-1)/(1+x^2)dx`

`= [tan^(-1)x(x^2/2)]_0^1-1/2int_0^1(1+x^2)/(1+x^2)dx+1/2int_0^1(1)/(1+x^2)dx`

`= [tan^(-1)x(x^2/2)]_0^1-1/2[x-tan^(-1)x]_0^1`

`= ((pi/4)xx1/2-0)-1/2(1-pi/4)`

`= pi/8+pi/8-1/2`

`= pi/4-1/2`

So the answer is;

`int_0^1 x tan^-1 x dx = pi/4-1/2`


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