What is the integral of xsqrt(3x+5) ?

The integral of xsqrt(3x+5) is 1/9 (2/5 (3x+5)^{5/2}-10/3 (3x+5)^{3/2})+C

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integral of xsqrt(3x+5)

\int x 3 x + 5 d x

\frac{1}{9} (\frac{2}{5} (3 x + 5)^{\frac{5}{2}} - \frac{10}{3} (3 x + 5)^{\frac{3}{2}}) + C

Solution steps

\int x 3 x + 5 d x

Apply u-substitution

: \int \frac{u ( u - 5 )}{9} d u

= \int \frac{u ( u - 5 )}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int u (u - 5) d u

Expand

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

= \frac{1}{9} \cdot \int u^{\frac{3}{2}} - 5 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}{9} (\int u^{\frac{3}{2}} d u - \int 5 u d u)

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

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

= \frac{1}{9} (\frac{2}{5} u^{\frac{5}{2}} - \frac{10}{3} u^{\frac{3}{2}})

Substitute back

u = 3 x + 5

= \frac{1}{9} (\frac{2}{5} (3 x + 5)^{\frac{5}{2}} - \frac{10}{3} (3 x + 5)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{9} (\frac{2}{5} (3 x + 5)^{\frac{5}{2}} - \frac{10}{3} (3 x + 5)^{\frac{3}{2}}) + C

\int sin^{4} (2 x) d x

\frac{\partial}{\partial x} (\frac{y}{( x ^{2} + y ^{2} )})

n = 1 \sum \infty arctan (5 n)

\frac{d}{d x} (I n (x^{2} + 9))

\frac{d}{d x} (e^{2 x} + e^{x})

What is the integral of xsqrt(3x+5) ?
The integral of xsqrt(3x+5) is 1/9 (2/5 (3x+5)^{5/2}-10/3 (3x+5)^{3/2})+C