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

The integral of x^2sqrt(2x+5) is 1/21 (3x^2-6x+10)(2x+5)^{3/2}+C

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integral of x^2sqrt(2x+5)

\int x^{2} 2 x + 5 d x

\frac{1}{21} (3 x^{2} - 6 x + 10) (2 x + 5)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 2 x + 5 d x

Apply u-substitution

: \int \frac{u ( u - 5 ) ^{2}}{8} d u

= \int \frac{u ( u - 5 ) ^{2}}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u (u - 5)^{2} d u

Expand

u (u - 5)^{2} : u^{\frac{5}{2}} - 10 u^{\frac{3}{2}} + 25 u

= \frac{1}{8} \cdot \int u^{\frac{5}{2}} - 10 u^{\frac{3}{2}} + 25 u 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}{8} (\int u^{\frac{5}{2}} d u - \int 10 u^{\frac{3}{2}} d u + \int 25 u d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 10 u^{\frac{3}{2}} d u = 4 u^{\frac{5}{2}}

\int 25 u d u = \frac{50}{3} u^{\frac{3}{2}}

= \frac{1}{8} (\frac{2}{7} u^{\frac{7}{2}} - 4 u^{\frac{5}{2}} + \frac{50}{3} u^{\frac{3}{2}})

Substitute back

u = 2 x + 5

= \frac{1}{8} (\frac{2}{7} (2 x + 5)^{\frac{7}{2}} - 4 (2 x + 5)^{\frac{5}{2}} + \frac{50}{3} (2 x + 5)^{\frac{3}{2}})

Simplify

\frac{1}{8} (\frac{2}{7} (2 x + 5)^{\frac{7}{2}} - 4 (2 x + 5)^{\frac{5}{2}} + \frac{50}{3} (2 x + 5)^{\frac{3}{2}}) : \frac{1}{21} (3 x^{2} - 6 x + 10) (2 x + 5)^{\frac{3}{2}}

= \frac{1}{21} (3 x^{2} - 6 x + 10) (2 x + 5)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{21} (3 x^{2} - 6 x + 10) (2 x + 5)^{\frac{3}{2}} + C

y = \frac{3}{x}, (4, \frac{3}{2})

t^{2} cos (t)

y^{^{''}} + 4 y^{^{'}} + 4 y = t + e^{2 t}

f (x) = 9 x^{2}

\int - 2 x^{3} + 12 x^{2} - 20 x + 8.5 d x

What is the integral of x^2sqrt(2x+5) ?
The integral of x^2sqrt(2x+5) is 1/21 (3x^2-6x+10)(2x+5)^{3/2}+C