# `int_1^2 x^4 (ln(x))^2 dx` Evaluate the integral

`int_1^2x^4(ln(x))^2dx`

If f(x) and g(x) are differentiable functions, then

`intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx`

If we write f(x)=u and ` ` g'(x)=v, then

`intuvdx=uintvdx-int(u'intvdx)dx`

Using the above method of integration by parts,

`intx^4(ln(x))^2dx=(ln(x))^2intx^4dx-int(d/dx(ln(x)^2)intx^4dx)dx`

`=(ln(x))^2*x^5/5-int(2ln(x)*1/x(x^5/5))dx`

`=(ln(x))^2x^5/5-2/5intx^4ln(x)dx`

again applying integration by parts,

`=x^5/5(ln(x))^2-2/5(ln(x)intx^4dx-int(d/dx(ln(x))intx^4dx)dx)`

`=x^5/5(ln(x))^2-2/5(ln(x)x^5/5-int(1/x*x^5/5)dx)`

`=x^5/5(ln(x))^2-2/25x^5ln(x)+2/25intx^4dx`

`=x^5/5(ln(x))^2-2/25x^5ln(x)+2/25*x^5/5`

`=x^5/5(ln(x))^2-2/25x^5ln(x)+2/125x^5+C`

Now evaluate the definite integral,

`int_1^2x^4(ln(x))^2dx=[x^5/5(ln(x))^2-2/25x^5ln(x)+2/125x^5]_1^2`

`=[2^5/5(ln(2))^2-2/25*2^5ln(2)+2/125(2^5)]-[1^5/5(ln(1))^2-2/25(1)^5ln(1)+2/125(1^5)]`

`=[32/5(ln(2))^2-64/25ln(2)+64/125]-[2/125]`

`=32/5(ln(2))^2-64/25ln(2)+62/125`

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