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

The integral of x(x+7)^4 is (x^6)/6+(28x^5)/5+(147x^4)/2+(1372x^3)/3+(2401x^2)/2+C

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integral of x(x+7)^4

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

\frac{x ^{6}}{6} + \frac{28 x ^{5}}{5} + \frac{147 x ^{4}}{2} + \frac{1372 x ^{3}}{3} + \frac{2401 x ^{2}}{2} + C

Solution steps

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

Expand

x (x + 7)^{4} : x^{5} + 28 x^{4} + 294 x^{3} + 1372 x^{2} + 2401 x

= \int x^{5} + 28 x^{4} + 294 x^{3} + 1372 x^{2} + 2401 x 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

= \int x^{5} d x + \int 28 x^{4} d x + \int 294 x^{3} d x + \int 1372 x^{2} d x + \int 2401 x d x

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

\int 28 x^{4} d x = \frac{28 x ^{5}}{5}

\int 294 x^{3} d x = \frac{147 x ^{4}}{2}

\int 1372 x^{2} d x = \frac{1372 x ^{3}}{3}

\int 2401 x d x = \frac{2401 x ^{2}}{2}

= \frac{x ^{6}}{6} + \frac{28 x ^{5}}{5} + \frac{147 x ^{4}}{2} + \frac{1372 x ^{3}}{3} + \frac{2401 x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{6}}{6} + \frac{28 x ^{5}}{5} + \frac{147 x ^{4}}{2} + \frac{1372 x ^{3}}{3} + \frac{2401 x ^{2}}{2} + C

(1 + x^{2}) \frac{d y}{d x} - x y = 0

\int_{2}^{5} 4 - 2 x d x

t \to a lim (ln (t))

\int t (t^{2} + 1)^{7} d t

\frac{d y}{d x} = e^{x}

What is the integral of x(x+7)^4 ?
The integral of x(x+7)^4 is (x^6)/6+(28x^5)/5+(147x^4)/2+(1372x^3)/3+(2401x^2)/2+C