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

The integral of xsqrt(2x+5) is 1/15 (3x-5)(2x+5)^{3/2}+C

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integral of xsqrt(2x+5)

\int x 2 x + 5 d x

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

Solution steps

\int x 2 x + 5 d x

Apply u-substitution

: \int \frac{u ( u - 5 )}{4} d u

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

Expand

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

= \frac{1}{4} \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}{4} (\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}{4} (\frac{2}{5} u^{\frac{5}{2}} - \frac{10}{3} u^{\frac{3}{2}})

Substitute back

u = 2 x + 5

= \frac{1}{4} (\frac{2}{5} (2 x + 5)^{\frac{5}{2}} - \frac{10}{3} (2 x + 5)^{\frac{3}{2}})

Simplify

\frac{1}{4} (\frac{2}{5} (2 x + 5)^{\frac{5}{2}} - \frac{10}{3} (2 x + 5)^{\frac{3}{2}}) : \frac{1}{15} (3 x - 5) (2 x + 5)^{\frac{3}{2}}

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

Add a constant to the solution

= \frac{1}{15} (3 x - 5) (2 x + 5)^{\frac{3}{2}} + C

f (x) = 25 x^{\frac{4}{5}}

\int tan^{2} (x) + tan^{4} (x) d x

n = 1 \sum \infty (- 5)^{n} x^{n}

arcsin (x^{2})

What is the integral of xsqrt(2x+5) ?
The integral of xsqrt(2x+5) is 1/15 (3x-5)(2x+5)^{3/2}+C