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

The integral of ((x^2))/(sqrt(2x+1)) is 1/15 (3x^2-2x+2)sqrt(2x+1)+C

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integral of ((x^2))/(sqrt(2x+1))

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

\frac{1}{15} (3 x^{2} - 2 x + 2) 2 x + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{( u ^{2} - 1 ) ^{2}}{4} d u

= \int \frac{( u ^{2} - 1 ) ^{2}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int (u^{2} - 1)^{2} d u

Expand

(u^{2} - 1)^{2} : u^{4} - 2 u^{2} + 1

= \frac{1}{4} \cdot \int u^{4} - 2 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{1}{4} (\int u^{4} d u - \int 2 u^{2} d u + \int 1 d u)

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

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

\int 1 d u = u

= \frac{1}{4} (\frac{u ^{5}}{5} - \frac{2 u ^{3}}{3} + u)

Substitute back

u = 2 x + 1

= \frac{1}{4} (\frac{( 2 x + 1 ) ^{5}}{5} - \frac{2 ( 2 x + 1 ) ^{3}}{3} + 2 x + 1)

Simplify

\frac{1}{4} (\frac{( 2 x + 1 ) ^{5}}{5} - \frac{2 ( 2 x + 1 ) ^{3}}{3} + 2 x + 1) : \frac{1}{15} (3 x^{2} - 2 x + 2) 2 x + 1

= \frac{1}{15} (3 x^{2} - 2 x + 2) 2 x + 1

Add a constant to the solution

= \frac{1}{15} (3 x^{2} - 2 x + 2) 2 x + 1 + C

\int \frac{x}{1 - x} d x

\frac{d}{d x} (5 (2 x cos (7 x) - 7 x^{2} sin (7 x)))

\int cos^{3} (20 x) d x

y^{^{''}} + 2 y = sin (2 x), y (0) = 0, y^{^{'}} (0) = 0

x \to 2 + lim (\frac{∣ x ∣}{2 - x})

What is the integral of ((x^2))/(sqrt(2x+1)) ?
The integral of ((x^2))/(sqrt(2x+1)) is 1/15 (3x^2-2x+2)sqrt(2x+1)+C