What is the integral of 14xsqrt(7x-3) ?

The integral of 14xsqrt(7x-3) is 2/7 (2/5 (7x-3)^{5/2}+2(7x-3)^{3/2})+C

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integral of 14xsqrt(7x-3)

\int 14 x 7 x - 3 d x

\frac{2}{7} (\frac{2}{5} (7 x - 3)^{\frac{5}{2}} + 2 (7 x - 3)^{\frac{3}{2}}) + C

Solution steps

\int 14 x 7 x - 3 d x

Take the constant out:

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

= 14 \cdot \int x 7 x - 3 d x

Apply u-substitution

: \int \frac{u ( u + 3 )}{49} d u

= 14 \cdot \int \frac{u ( u + 3 )}{49} d u

Take the constant out:

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

= 14 \cdot \frac{1}{49} \cdot \int u (u + 3) d u

Expand

u (u + 3) : u^{\frac{3}{2}} + 3 u

= 14 \cdot \frac{1}{49} \cdot \int u^{\frac{3}{2}} + 3 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

= 14 \cdot \frac{1}{49} (\int u^{\frac{3}{2}} d u + \int 3 u d u)

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

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

= 14 \cdot \frac{1}{49} (\frac{2}{5} u^{\frac{5}{2}} + 2 u^{\frac{3}{2}})

Substitute back

u = 7 x - 3

= 14 \cdot \frac{1}{49} (\frac{2}{5} (7 x - 3)^{\frac{5}{2}} + 2 (7 x - 3)^{\frac{3}{2}})

Simplify

14 \cdot \frac{1}{49} (\frac{2}{5} (7 x - 3)^{\frac{5}{2}} + 2 (7 x - 3)^{\frac{3}{2}}) : \frac{2}{7} (\frac{2}{5} (7 x - 3)^{\frac{5}{2}} + 2 (7 x - 3)^{\frac{3}{2}})

= \frac{2}{7} (\frac{2}{5} (7 x - 3)^{\frac{5}{2}} + 2 (7 x - 3)^{\frac{3}{2}})

Add a constant to the solution

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

\frac{d}{d x} (e^{5 x s i n (2 x)})

\int 5 cos (x) + 3 sin (x) d x

f (x) = 2 x^{2} + 8 x

n = 1 \sum \infty 2 (- \frac{1}{2})^{n}

\frac{d}{d t} (1 + t)

What is the integral of 14xsqrt(7x-3) ?
The integral of 14xsqrt(7x-3) is 2/7 (2/5 (7x-3)^{5/2}+2(7x-3)^{3/2})+C