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

The integral of x^2sqrt(7x-5) is 1/343 (2/7 (7x-5)^{7/2}+4(7x-5)^{5/2}+50/3 (7x-5)^{3/2})+C

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

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

\frac{1}{343} (\frac{2}{7} (7 x - 5)^{\frac{7}{2}} + 4 (7 x - 5)^{\frac{5}{2}} + \frac{50}{3} (7 x - 5)^{\frac{3}{2}}) + C

Solution steps

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

Apply u-substitution

: \int \frac{u ( u + 5 ) ^{2}}{343} d u

= \int \frac{u ( u + 5 ) ^{2}}{343} d u

Take the constant out:

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

= \frac{1}{343} \cdot \int u (u + 5)^{2} d u

Expand

u (u + 5)^{2} : u^{\frac{5}{2}} + 10 u^{\frac{3}{2}} + 25 u

= \frac{1}{343} \cdot \int u^{\frac{5}{2}} + 10 u^{\frac{3}{2}} + 25 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}{343} (\int u^{\frac{5}{2}} d u + \int 10 u^{\frac{3}{2}} d u + \int 25 u d u)

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

\int 10 u^{\frac{3}{2}} d u = 4 u^{\frac{5}{2}}

\int 25 u d u = \frac{50}{3} u^{\frac{3}{2}}

= \frac{1}{343} (\frac{2}{7} u^{\frac{7}{2}} + 4 u^{\frac{5}{2}} + \frac{50}{3} u^{\frac{3}{2}})

Substitute back

u = 7 x - 5

= \frac{1}{343} (\frac{2}{7} (7 x - 5)^{\frac{7}{2}} + 4 (7 x - 5)^{\frac{5}{2}} + \frac{50}{3} (7 x - 5)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{343} (\frac{2}{7} (7 x - 5)^{\frac{7}{2}} + 4 (7 x - 5)^{\frac{5}{2}} + \frac{50}{3} (7 x - 5)^{\frac{3}{2}}) + C

t^{3} \frac{d y}{d t} + 4 t^{2} y = e^{- t}

\int \frac{t t + t}{t ^{2}} d t

R (a) = (3 a + 1)^{2}

y^{^{'}} + 4 y = t^{2} e^{- 4 t}

\int \frac{2 x + 7}{9 - 4 x ^{2}} d x

What is the integral of x^2sqrt(7x-5) ?
The integral of x^2sqrt(7x-5) is 1/343 (2/7 (7x-5)^{7/2}+4(7x-5)^{5/2}+50/3 (7x-5)^{3/2})+C