# Use integration by parts to find the indefinite integral. `int sqrt (2x) ln x^2 dx`

lemjay | High School Teacher | (Level 2) Senior Educator

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`int sqrt(2x) ln x^2 dx`

The formula of integration by parts is `intudv = uv - vdu` .

So let,

`u= ln x^2 = 2lnx `         and        `dv = sqrt(2x)dx = (2x)^(1/2)dx`

`du =2/xdx`                                     `v = int (2x)^(1/2) dx`

To solve for v, use substitution method. Let,

`y= 2x`            `dy = 2dx`            `(dy)/2 =dx`

So we have,

`v= int (2x)^(1/2) dx = int y^(1/2) (dy)/2 = 1/2 int y^(1/2)dy = 1/2* y^(3/2)/(3/2)= y^(3/2)/3`

`v = (2x)^(3/2)/3 = [2^(3/2)x^(3/2)]/3`

Substitute u, v, and dv to the formula of integration by parts.

`int sqrt(2x)lnx^2 = 2lnx*[2^(3/2)x^(3/2)]/3 - int [2^(3/2)x^(3/2)]/3 2/xdx`

`=[2^(5/2) x^(3/2)lnx]/3 - 2^(5/2)/3 int x^1/2 dx`

`= [2^(5/2)x^(3/2)lnx]/3-[2^(7/2)x^(3/2)]/9 + C`

`=[2^(5/2)x^(3/2)]/3 (lnx - 2/3) + C`

`= (4xsqrt(2x))/3(lnx-2/3) + C`

Hence, `int sqrt(2x) lnx^2 dx = [4xsqrt(2x)]/3(lnx-2/3)+ C` .