What is the integral of (3x)/(sqrt(3-7x)) ?

The integral of (3x)/(sqrt(3-7x)) is 6/49 (1/3 (3-7x)^{3/2}-3sqrt(3-7x))+C

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integral of (3x)/(sqrt(3-7x))

\int \frac{3 x}{3 - 7 x} d x

\frac{6}{49} (\frac{1}{3} (3 - 7 x)^{\frac{3}{2}} - 3 3 - 7 x) + C

Solution steps

\int \frac{3 x}{3 - 7 x} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{3 - 7 x} d x

Apply u-substitution

= 3 \cdot \int \frac{2 ( u ^{2} - 3 )}{49} d u

Take the constant out:

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

= 3 \cdot \frac{2}{49} \cdot \int 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

= 3 \cdot \frac{2}{49} (\int u^{2} d u - \int 3 d u)

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

\int 3 d u = 3 u

= 3 \cdot \frac{2}{49} (\frac{u ^{3}}{3} - 3 u)

Substitute back

u = 3 - 7 x

= 3 \cdot \frac{2}{49} (\frac{( 3 - 7 x ) ^{3}}{3} - 3 3 - 7 x)

Simplify

3 \cdot \frac{2}{49} (\frac{( 3 - 7 x ) ^{3}}{3} - 3 3 - 7 x) : \frac{6}{49} (\frac{1}{3} (3 - 7 x)^{\frac{3}{2}} - 3 3 - 7 x)

= \frac{6}{49} (\frac{1}{3} (3 - 7 x)^{\frac{3}{2}} - 3 3 - 7 x)

Add a constant to the solution

= \frac{6}{49} (\frac{1}{3} (3 - 7 x)^{\frac{3}{2}} - 3 3 - 7 x) + C

What is the integral of (3x)/(sqrt(3-7x)) ?
The integral of (3x)/(sqrt(3-7x)) is 6/49 (1/3 (3-7x)^{3/2}-3sqrt(3-7x))+C