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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Substitute back

u = a + b x

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

Simplify

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

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

Add a constant to the solution

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

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