Question

integration of (1+x-1/x)e^x+1/xdx

Answer 1

Let, I=∫(1+x−1x)⋅ex+1xdxLet, I=∫(1+x−1x)⋅ex+1xdx

=∫(1+x(1−1x2))⋅ex+1xdx=∫(1+x(1−1x2))⋅ex+1xdx

=∫ex+1x⋅dx+∫x1⋅ex+1x(1−1x2)2dx=∫ex+1x⋅dx+∫x⏟1⋅ex+1x(1−1x2)⏟2dx

=∫ex+1x⋅dx+xex+1x−∫d(x)dx⋅ex+1x⋅dx+c;where, c is constant of integration.=∫ex+1x⋅dx+xex+1x−∫d(x)dx⋅ex+1x⋅dx+c;where, c is constant of integration.

[Integrating by-parts(second integral) &[Integrating by-parts(second integral) &

since, ∫ex+1x⋅(1−1x2)dxsince, ∫ex+1x⋅(1−1x2)dx

=∫ex+1x⋅d(x+1x)=∫ex+1x⋅d(x+1x)

=ex+1x.]=ex+1x.]

=∫ex+1x⋅dx+xex+1x−∫ex+1x⋅dx+c=∫ex+1x⋅dx+xex+1x−∫ex+1x⋅dx+c

∴I=xex+1x+c.†∴I=xex+1x+c.†

Aliter : –––––––––Aliter : _

We have,

I=∫(e1x+xe1x−1xe1x)exdxI=∫(e1x+xe1x−1xe1x)exdx

=∫ex{(xe1x)+(e1x−1xe1x)}dx=∫ex{(xe1x)+(e1x−1xe1x)}dx

=∫ex{f(x)+f′(x)}dx;where, f(x)=xe1x.=∫ex{f(x)+f′(x)}dx;where, f(x)=xe1x.

=exf(x)+c=exf(x)+c

=ex⋅xe1x+c=ex⋅xe1x+c

∴I=xex+1x+c.†∴I=xex+1x+c.†

Hope, you'll understand..!!

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Thank You!