What is the integral of ((b/2)x)/(2sqrt(a+bx)) ?

The integral of ((b/2)x)/(2sqrt(a+bx)) is ((bx+a)^{3/2}-3asqrt(bx+a))/(6b)+C

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integral of ((b/2)x)/(2sqrt(a+bx))

\int \frac{( \frac{b}{2} ) x}{2 a + b x} d x

\frac{( b x + a ) ^{\frac{3}{2}} - 3 a b x + a}{6 b} + C

Solution steps

\int \frac{\frac{b}{2} x}{2 a + b x} d x

Take the constant out:

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

= \frac{\frac{b}{2}}{2} \cdot \int \frac{x}{a + b x} d x

Simplify

\frac{\frac{b}{2}}{2} : \frac{b}{4}

= \frac{b}{4} \cdot \int \frac{x}{a + b x} d x

Apply u-substitution

: \int \frac{2 ( u ^{2} - a )}{b ^{2}} d u

= \frac{b}{4} \cdot \int \frac{2 ( u ^{2} - a )}{b ^{2}} d u

Take the constant out:

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

= \frac{b}{4} \cdot \frac{2}{b ^{2}} \cdot \int u^{2} - a 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{b}{4} \cdot \frac{2}{b ^{2}} (\int u^{2} d u - \int a d u)

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

\int a d u = a u

= \frac{b}{4} \cdot \frac{2}{b ^{2}} (\frac{u ^{3}}{3} - a u)

Substitute back

u = a + b x

= \frac{b}{4} \cdot \frac{2}{b ^{2}} (\frac{( a + b x ) ^{3}}{3} - a a + b x)

Simplify

\frac{b}{4} \cdot \frac{2}{b ^{2}} (\frac{( a + b x ) ^{3}}{3} - a a + b x) : \frac{( b x + a ) ^{\frac{3}{2}} - 3 a b x + a}{6 b}

= \frac{( b x + a ) ^{\frac{3}{2}} - 3 a b x + a}{6 b}

Add a constant to the solution

= \frac{( b x + a ) ^{\frac{3}{2}} - 3 a b x + a}{6 b} + C

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What is the integral of ((b/2)x)/(2sqrt(a+bx)) ?
The integral of ((b/2)x)/(2sqrt(a+bx)) is ((bx+a)^{3/2}-3asqrt(bx+a))/(6b)+C