The integral ∫ 1 2 e x .   x x   2 + log e   x   dx equals :

Question

The integral &#8747; 1 2 e x . &#160; x x &#160; 2 + log e &#160; x &#160; dx equals :

Aicharya · Accepted Answer

Let y = ex x l ny = x 1 + l nx 1 y dy dx = 2 + l nx &#8658; dy = ex x 2 + l nx &#160; dx &#8747; 1 2 e x . x x 2 + log e x dx = y 1 2 = ex x 1 2 = 4 e 2 &#8722; e

The integral &#8747; 1 2 e x . &#160; x x &#160; 2 + log e &#160; x &#160; dx equals :

Solution

The integral ∫ 1 2 e x . x x 2 + log e x dx equals :