What is the integral of x/(sqrt(ax+b)) ?

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

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integral of x/(sqrt(ax+b))

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

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

\int b d u = b u

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

Substitute back

u = a x + b

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

Simplify

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

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

Add a constant to the solution

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

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