What is the integral of s/(sqrt(2s+3)) ?

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

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integral of s/(sqrt(2s+3))

\int \frac{s}{2 s + 3} d s

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

Solution steps

\int \frac{s}{2 s + 3} d s

Apply u-substitution

: \int \frac{u ^{2} - 3}{2} d u

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

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

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

\int 3 d u = 3 u

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

Substitute back

u = 2 s + 3

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

Simplify

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

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

Add a constant to the solution

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

f (x) = 6 + 5 x^{2} - 2 x^{3}, at x = 2

\int x (3 + 5 x) d x

x \to 2 lim (3 f (x) + 2 g (x) (x)^{2})

\frac{2 s ^{2} + 5 s + 5}{( s + 1 ) ^{3}}

(\frac{5}{x}), at x = 4

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