What is the integral from 3/2 to 6 of x/(sqrt(2x-3)) ?

The integral from 3/2 to 6 of x/(sqrt(2x-3)) is 9

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integral from 3/2 to 6 of x/(sqrt(2x-3))

\int_{\frac{3}{2}}^{6} \frac{x}{2 x - 3} d x

9

Solution steps

\int_{\frac{3}{2}}^{6} \frac{x}{2 x - 3} d x

Apply u-substitution

= \int_{0}^{3} \frac{u ^{2} + 3}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{3} u^{2} + 3 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{3} u^{2} d u + \int_{0}^{3} 3 d u)

\int_{0}^{3} u^{2} d u = 9

\int_{0}^{3} 3 d u = 9

= \frac{1}{2} (9 + 9)

Simplify

\frac{1}{2} (9 + 9) : 9

= 9

What is the integral from 3/2 to 6 of x/(sqrt(2x-3)) ?
The integral from 3/2 to 6 of x/(sqrt(2x-3)) is 9