What is the integral of 30(x+5)(x+7)^4 ?

The integral of 30(x+5)(x+7)^4 is 30((x^6)/6+(33x^5)/5+(217x^4)/2+(2842x^3)/3+(9261x^2)/2+12005x)+C

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integral of 30(x+5)(x+7)^4

\int 30 (x + 5) (x + 7)^{4} d x

30 (\frac{x ^{6}}{6} + \frac{33 x ^{5}}{5} + \frac{217 x ^{4}}{2} + \frac{2842 x ^{3}}{3} + \frac{9261 x ^{2}}{2} + 12005 x) + C

Solution steps

\int 30 (x + 5) (x + 7)^{4} d x

Take the constant out:

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

= 30 \cdot \int (x + 5) (x + 7)^{4} d x

Expand

(x + 5) (x + 7)^{4} : x^{5} + 33 x^{4} + 434 x^{3} + 2842 x^{2} + 9261 x + 12005

= 30 \cdot \int x^{5} + 33 x^{4} + 434 x^{3} + 2842 x^{2} + 9261 x + 12005 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 30 (\int x^{5} d x + \int 33 x^{4} d x + \int 434 x^{3} d x + \int 2842 x^{2} d x + \int 9261 x d x + \int 12005 d x)

\int x^{5} d x = \frac{x ^{6}}{6}

\int 33 x^{4} d x = \frac{33 x ^{5}}{5}

\int 434 x^{3} d x = \frac{217 x ^{4}}{2}

\int 2842 x^{2} d x = \frac{2842 x ^{3}}{3}

\int 9261 x d x = \frac{9261 x ^{2}}{2}

\int 12005 d x = 12005 x

= 30 (\frac{x ^{6}}{6} + \frac{33 x ^{5}}{5} + \frac{217 x ^{4}}{2} + \frac{2842 x ^{3}}{3} + \frac{9261 x ^{2}}{2} + 12005 x)

Add a constant to the solution

= 30 (\frac{x ^{6}}{6} + \frac{33 x ^{5}}{5} + \frac{217 x ^{4}}{2} + \frac{2842 x ^{3}}{3} + \frac{9261 x ^{2}}{2} + 12005 x) + C

y^{^{'}} = x (5 - y)

d er i v a t i v e \frac{g ( x )}{1 + f ( x )}

x \to - 3 lim (\frac{3 x}{9 - x ^{2}})

\frac{d y}{d x} = \frac{( x - 1 )}{y + 2}

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

What is the integral of 30(x+5)(x+7)^4 ?
The integral of 30(x+5)(x+7)^4 is 30((x^6)/6+(33x^5)/5+(217x^4)/2+(2842x^3)/3+(9261x^2)/2+12005x)+C