What is the integral of (x+3)(x-1)^5 ?

The integral of (x+3)(x-1)^5 is (x^7)/7-(x^6)/3-x^5+5x^4-(25x^3)/3+7x^2-3x+C

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integral of (x+3)(x-1)^5

\int (x + 3) (x - 1)^{5} d x

\frac{x ^{7}}{7} - \frac{x ^{6}}{3} - x^{5} + 5 x^{4} - \frac{25 x ^{3}}{3} + 7 x^{2} - 3 x + C

Solution steps

\int (x + 3) (x - 1)^{5} d x

Expand

(x + 3) (x - 1)^{5} : x^{6} - 2 x^{5} - 5 x^{4} + 20 x^{3} - 25 x^{2} + 14 x - 3

= \int x^{6} - 2 x^{5} - 5 x^{4} + 20 x^{3} - 25 x^{2} + 14 x - 3 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^{6} d x - \int 2 x^{5} d x - \int 5 x^{4} d x + \int 20 x^{3} d x - \int 25 x^{2} d x + \int 14 x d x - \int 3 d x

\int x^{6} d x = \frac{x ^{7}}{7}

\int 2 x^{5} d x = \frac{x ^{6}}{3}

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

\int 20 x^{3} d x = 5 x^{4}

\int 25 x^{2} d x = \frac{25 x ^{3}}{3}

\int 14 x d x = 7 x^{2}

\int 3 d x = 3 x

= \frac{x ^{7}}{7} - \frac{x ^{6}}{3} - x^{5} + 5 x^{4} - \frac{25 x ^{3}}{3} + 7 x^{2} - 3 x

Add a constant to the solution

= \frac{x ^{7}}{7} - \frac{x ^{6}}{3} - x^{5} + 5 x^{4} - \frac{25 x ^{3}}{3} + 7 x^{2} - 3 x + C

\frac{d x}{d t} = 4 x (7 - x)

f (x) = \frac{4}{3 x - 1}, at x = 1

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

f (x) = tan (x) \cdot cot (x)

\int \frac{8}{( x - 4 ) ^{2}} d x

What is the integral of (x+3)(x-1)^5 ?
The integral of (x+3)(x-1)^5 is (x^7)/7-(x^6)/3-x^5+5x^4-(25x^3)/3+7x^2-3x+C