What is the integral of (2x+5)/(sqrt(2x+6)) ?

The integral of (2x+5)/(sqrt(2x+6)) is 1/3 (2x+6)^{3/2}-sqrt(2x+6)+C

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integral of (2x+5)/(sqrt(2x+6))

\int \frac{2 x + 5}{2 x + 6} d x

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

Solution steps

\int \frac{2 x + 5}{2 x + 6} d x

Apply u-substitution

: \int u^{2} - 1 d u

= \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

= \int u^{2} d u - \int 1 d u

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

\int 1 d u = u

= \frac{u ^{3}}{3} - u

Substitute back

u = 2 x + 6

= \frac{( 2 x + 6 ) ^{3}}{3} - 2 x + 6

Simplify

\frac{( 2 x + 6 ) ^{3}}{3} - 2 x + 6 : \frac{1}{3} (2 x + 6)^{\frac{3}{2}} - 2 x + 6

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{x ^{3}}{3} + \frac{x ^{2}}{2} + e^{- x})

(1 + x^{2})^{- 1}

(- 1.1) (1.6)

f (x) = x^{6}

\int \frac{x + 1}{x ^{2} + 4 x + 7} d x

What is the integral of (2x+5)/(sqrt(2x+6)) ?
The integral of (2x+5)/(sqrt(2x+6)) is 1/3 (2x+6)^{3/2}-sqrt(2x+6)+C