What is the integral of x/(sqrt(7x+1)) ?

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

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integral of x/(sqrt(7x+1))

\int \frac{x}{7 x + 1} d x

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

Solution steps

\int \frac{x}{7 x + 1} d x

Apply u-substitution

= \int \frac{2 ( u ^{2} - 1 )}{49} d u

Take the constant out:

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

= \frac{2}{49} \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{2}{49} (\int u^{2} d u - \int 1 d u)

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

\int 1 d u = u

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

Substitute back

u = 7 x + 1

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

Simplify

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

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

Add a constant to the solution

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

d er i v a t i v e f (x) = 9 - 2 x^{2}

\int \frac{1}{2 x + 2 x} d x

\frac{d}{d x} (arccos (x^{2} + y^{2} - 4))

y = c e^{x}

y^{^{''}} - 2 y^{^{'}} + 8 y = 0

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