What is the integral of (5x+1)^{1/2} ?

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

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integral of (5x+1)^{1/2}

\int (5 x + 1)^{\frac{1}{2}} d x

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

Solution steps

\int (5 x + 1)^{\frac{1}{2}} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{2}}}{5} d u

= \int \frac{u ^{\frac{1}{2}}}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int u^{\frac{1}{2}} d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{5} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 5 x + 1

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + 4 y = 3 cos (2 t)

\frac{d}{d n} (n^{n})

n = 0 \sum \infty n! (x - 5)^{n}

\int \frac{1}{x ( 1 + ln ( x ) ) ^{3}} d x

\frac{d r}{d θ} = \frac{r ^{2} + 2 r θ}{θ ^{2}}

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