What is the integral of (x+1)(x+3)^2 ?

The integral of (x+1)(x+3)^2 is (x^4)/4+(7x^3)/3+(15x^2)/2+9x+C

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integral of (x+1)(x+3)^2

\int (x + 1) (x + 3)^{2} d x

\frac{x ^{4}}{4} + \frac{7 x ^{3}}{3} + \frac{15 x ^{2}}{2} + 9 x + C

Solution steps

\int (x + 1) (x + 3)^{2} d x

Expand

(x + 1) (x + 3)^{2} : x^{3} + 7 x^{2} + 15 x + 9

= \int x^{3} + 7 x^{2} + 15 x + 9 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^{3} d x + \int 7 x^{2} d x + \int 15 x d x + \int 9 d x

\int x^{3} d x = \frac{x ^{4}}{4}

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

\int 15 x d x = \frac{15 x ^{2}}{2}

\int 9 d x = 9 x

= \frac{x ^{4}}{4} + \frac{7 x ^{3}}{3} + \frac{15 x ^{2}}{2} + 9 x

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{7 x ^{3}}{3} + \frac{15 x ^{2}}{2} + 9 x + C

\frac{d}{d x} (\frac{32}{x})

- x^{2} + 9, 1.25 x + 3

y^{^{'}} - y = 9 t e^{2 t}, y (0) = 1

x \to 0 - lim (\frac{1}{x \cdot e ^{\frac{1}{x}}})

\frac{4}{s + 4}

What is the integral of (x+1)(x+3)^2 ?
The integral of (x+1)(x+3)^2 is (x^4)/4+(7x^3)/3+(15x^2)/2+9x+C