What is the integral of (2x+3)sqrt(2x+1) ?

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

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integral of (2x+3)sqrt(2x+1)

\int (2 x + 3) 2 x + 1 d x

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

Solution steps

\int (2 x + 3) 2 x + 1 d x

Apply u-substitution

: \int u^{2} (u^{2} + 2) d u

= \int u^{2} (u^{2} + 2) d u

Expand

u^{2} (u^{2} + 2) : u^{4} + 2 u^{2}

= \int u^{4} + 2 u^{2} 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^{4} d u + \int 2 u^{2} d u

\int u^{4} d u = \frac{u ^{5}}{5}

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

= \frac{u ^{5}}{5} + \frac{2 u ^{3}}{3}

Substitute back

u = 2 x + 1

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{5} (2 x + 1)^{\frac{5}{2}} + \frac{2}{3} (2 x + 1)^{\frac{3}{2}} + C

\frac{d}{d x} (sech^{2} (3 x))

\int \frac{2 x - 3}{x ^{2} + 4 x + 5} d x

n = 2 \sum \infty \frac{- 1}{n}

\frac{d}{d x} (\frac{x ^{3} - 2 x ^{2} + 2 x - 4}{x ^{2} + 2})

f (x) = \frac{c ^{9} - x ^{9}}{c ^{9} + x ^{9}}

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