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

The integral of (x+1)(x+3) is (x^3)/3+2x^2+3x+C

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

\int (x + 1) (x + 3) d x

\frac{x ^{3}}{3} + 2 x^{2} + 3 x + C

Solution steps

\int (x + 1) (x + 3) d x

Expand

(x + 1) (x + 3) : x^{2} + 4 x + 3

= \int x^{2} + 4 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^{2} d x + \int 4 x d x + \int 3 d x

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

\int 4 x d x = 2 x^{2}

\int 3 d x = 3 x

= \frac{x ^{3}}{3} + 2 x^{2} + 3 x

Add a constant to the solution

= \frac{x ^{3}}{3} + 2 x^{2} + 3 x + C

\frac{- 0.1 s}{s ^{2} - 9.8}

y = \frac{4}{3} (x + 1)^{\frac{3}{2}}

\int_{0}^{- 0.5} 150 x d x

(x^{2} + 5)^{2} (x - 5)

\int 6 x^{2} + 16 x - 10 d x

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