What is the integral of 2xsqrt(5x+1) ?

The integral of 2xsqrt(5x+1) is 2/25 (2/5 (5x+1)^{5/2}-2/3 (5x+1)^{3/2})+C

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integral of 2xsqrt(5x+1)

\int 2 x 5 x + 1 d x

\frac{2}{25} (\frac{2}{5} (5 x + 1)^{\frac{5}{2}} - \frac{2}{3} (5 x + 1)^{\frac{3}{2}}) + C

Solution steps

\int 2 x 5 x + 1 d x

Take the constant out:

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

= 2 \cdot \int x 5 x + 1 d x

Apply u-substitution

: \int \frac{u ( u - 1 )}{25} d u

= 2 \cdot \int \frac{u ( u - 1 )}{25} d u

Take the constant out:

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

= 2 \cdot \frac{1}{25} \cdot \int u (u - 1) d u

Expand

u (u - 1) : u^{\frac{3}{2}} - u

= 2 \cdot \frac{1}{25} \cdot \int u^{\frac{3}{2}} - 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

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

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

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

= 2 \cdot \frac{1}{25} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 5 x + 1

= 2 \cdot \frac{1}{25} (\frac{2}{5} (5 x + 1)^{\frac{5}{2}} - \frac{2}{3} (5 x + 1)^{\frac{3}{2}})

Simplify

2 \cdot \frac{1}{25} (\frac{2}{5} (5 x + 1)^{\frac{5}{2}} - \frac{2}{3} (5 x + 1)^{\frac{3}{2}}) : \frac{2}{25} (\frac{2}{5} (5 x + 1)^{\frac{5}{2}} - \frac{2}{3} (5 x + 1)^{\frac{3}{2}})

= \frac{2}{25} (\frac{2}{5} (5 x + 1)^{\frac{5}{2}} - \frac{2}{3} (5 x + 1)^{\frac{3}{2}})

Add a constant to the solution

= \frac{2}{25} (\frac{2}{5} (5 x + 1)^{\frac{5}{2}} - \frac{2}{3} (5 x + 1)^{\frac{3}{2}}) + C

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What is the integral of 2xsqrt(5x+1) ?
The integral of 2xsqrt(5x+1) is 2/25 (2/5 (5x+1)^{5/2}-2/3 (5x+1)^{3/2})+C