What is the integral of x/(sqrt(2x+7)) ?

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

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integral of x/(sqrt(2x+7))

\int \frac{x}{2 x + 7} d x

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

Solution steps

\int \frac{x}{2 x + 7} d x

Apply u-substitution

: \int \frac{u ^{2} - 7}{2} d u

= \int \frac{u ^{2} - 7}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{2} - 7 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{1}{2} (\int u^{2} d u - \int 7 d u)

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

\int 7 d u = 7 u

= \frac{1}{2} (\frac{u ^{3}}{3} - 7 u)

Substitute back

u = 2 x + 7

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (5 cos (7 x - 3 y))

x \to 2 + lim (\frac{1}{∣ 2 - x ∣})

f (x) = (3 x^{6} + 8 x^{3})^{4}

\frac{d}{d x} (\frac{1}{x ^{2} + a ^{2}})

\frac{5}{s ^{8}}

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