What is the integral of x^2sqrt(5x+7) ?

The integral of x^2sqrt(5x+7) is 1/125 (2/7 (5x+7)^{7/2}-28/5 (5x+7)^{5/2}+98/3 (5x+7)^{3/2})+C

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integral of x^2sqrt(5x+7)

\int x^{2} 5 x + 7 d x

\frac{1}{125} (\frac{2}{7} (5 x + 7)^{\frac{7}{2}} - \frac{28}{5} (5 x + 7)^{\frac{5}{2}} + \frac{98}{3} (5 x + 7)^{\frac{3}{2}}) + C

Solution steps

\int x^{2} 5 x + 7 d x

Apply u-substitution

: \int \frac{u ( u - 7 ) ^{2}}{125} d u

= \int \frac{u ( u - 7 ) ^{2}}{125} d u

Take the constant out:

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

= \frac{1}{125} \cdot \int u (u - 7)^{2} d u

Expand

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

= \frac{1}{125} \cdot \int u^{\frac{5}{2}} - 14 u^{\frac{3}{2}} + 49 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}{125} (\int u^{\frac{5}{2}} d u - \int 14 u^{\frac{3}{2}} d u + \int 49 u d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 14 u^{\frac{3}{2}} d u = \frac{28}{5} u^{\frac{5}{2}}

\int 49 u d u = \frac{98}{3} u^{\frac{3}{2}}

= \frac{1}{125} (\frac{2}{7} u^{\frac{7}{2}} - \frac{28}{5} u^{\frac{5}{2}} + \frac{98}{3} u^{\frac{3}{2}})

Substitute back

u = 5 x + 7

= \frac{1}{125} (\frac{2}{7} (5 x + 7)^{\frac{7}{2}} - \frac{28}{5} (5 x + 7)^{\frac{5}{2}} + \frac{98}{3} (5 x + 7)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{125} (\frac{2}{7} (5 x + 7)^{\frac{7}{2}} - \frac{28}{5} (5 x + 7)^{\frac{5}{2}} + \frac{98}{3} (5 x + 7)^{\frac{3}{2}}) + C

\frac{d}{d x} (\frac{6 x}{x ^{2} + 9})

\frac{\partial}{\partial y} (\frac{a x + b y}{c x + d y})

\frac{8}{9 t ^{6}}

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

x \to - 1 lim (x^{5})

What is the integral of x^2sqrt(5x+7) ?
The integral of x^2sqrt(5x+7) is 1/125 (2/7 (5x+7)^{7/2}-28/5 (5x+7)^{5/2}+98/3 (5x+7)^{3/2})+C