What is the integral of x^2sqrt(3x+5) ?

The integral of x^2sqrt(3x+5) is 1/27 (2/7 (3x+5)^{7/2}-4(3x+5)^{5/2}+50/3 (3x+5)^{3/2})+C

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integral of x^2sqrt(3x+5)

\int x^{2} 3 x + 5 d x

\frac{1}{27} (\frac{2}{7} (3 x + 5)^{\frac{7}{2}} - 4 (3 x + 5)^{\frac{5}{2}} + \frac{50}{3} (3 x + 5)^{\frac{3}{2}}) + C

Solution steps

\int x^{2} 3 x + 5 d x

Apply u-substitution

: \int \frac{u ( u - 5 ) ^{2}}{27} d u

= \int \frac{u ( u - 5 ) ^{2}}{27} d u

Take the constant out:

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

= \frac{1}{27} \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}{27} \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}{27} (\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}{27} (\frac{2}{7} u^{\frac{7}{2}} - 4 u^{\frac{5}{2}} + \frac{50}{3} u^{\frac{3}{2}})

Substitute back

u = 3 x + 5

= \frac{1}{27} (\frac{2}{7} (3 x + 5)^{\frac{7}{2}} - 4 (3 x + 5)^{\frac{5}{2}} + \frac{50}{3} (3 x + 5)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{27} (\frac{2}{7} (3 x + 5)^{\frac{7}{2}} - 4 (3 x + 5)^{\frac{5}{2}} + \frac{50}{3} (3 x + 5)^{\frac{3}{2}}) + C

f (x) = sin (3 x + 2)

\int_{1}^{5} 2 x - 1 d x

\int \frac{1}{4 + x} d x

\frac{d y}{d t} = t^{2} y^{3}, y (0) = - 1

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

What is the integral of x^2sqrt(3x+5) ?
The integral of x^2sqrt(3x+5) is 1/27 (2/7 (3x+5)^{7/2}-4(3x+5)^{5/2}+50/3 (3x+5)^{3/2})+C