What is the integral of (3x)/(2sqrt(x-1)) ?

The integral of (3x)/(2sqrt(x-1)) is (x-1)^{3/2}+3sqrt(x-1)+C

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integral of (3x)/(2sqrt(x-1))

\int \frac{3 x}{2 x - 1} d x

(x - 1)^{\frac{3}{2}} + 3 x - 1 + C

Solution steps

\int \frac{3 x}{2 x - 1} d x

Take the constant out:

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

= \frac{3}{2} \cdot \int \frac{x}{x - 1} d x

Apply u-substitution

: \int 2 (u^{2} + 1) d u

= \frac{3}{2} \cdot \int 2 (u^{2} + 1) d u

Take the constant out:

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

= \frac{3}{2} \cdot 2 \cdot \int u^{2} + 1 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{3}{2} \cdot 2 (\int u^{2} d u + \int 1 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

= \frac{3}{2} \cdot 2 (\frac{u ^{3}}{3} + u)

Substitute back

u = x - 1

= \frac{3}{2} \cdot 2 (\frac{( x - 1 ) ^{3}}{3} + x - 1)

Simplify

\frac{3}{2} \cdot 2 (\frac{( x - 1 ) ^{3}}{3} + x - 1) : (x - 1)^{\frac{3}{2}} + 3 x - 1

= (x - 1)^{\frac{3}{2}} + 3 x - 1

Add a constant to the solution

= (x - 1)^{\frac{3}{2}} + 3 x - 1 + C

2^{2 x}

\int x 4 x^{2} + 1 d x

\frac{d}{d x} (e^{5 x - 4 x^{2}})

\frac{d}{d x} (\frac{x}{7 x + 1})

1 - x^{2}, x - 1

What is the integral of (3x)/(2sqrt(x-1)) ?
The integral of (3x)/(2sqrt(x-1)) is (x-1)^{3/2}+3sqrt(x-1)+C