What is the integral from 0 to 4 of 6x+1/((x+3)^2) ?

The integral from 0 to 4 of 6x+1/((x+3)^2) is 1012/21

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integral from 0 to 4 of 6x+1/((x+3)^2)

\int_{0}^{4} 6 x + \frac{1}{( x + 3 ) ^{2}} d x

\frac{1012}{21}

Decimal Notation

48.19047 \dots

Solution steps

\int_{0}^{4} 6 x + \frac{1}{( x + 3 ) ^{2}} 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_{0}^{4} 6 x d x + \int_{0}^{4} \frac{1}{( x + 3 ) ^{2}} d x

\int_{0}^{4} 6 x d x = 48

\int_{0}^{4} \frac{1}{( x + 3 ) ^{2}} d x = \frac{4}{21}

= 48 + \frac{4}{21}

Simplify

48 + \frac{4}{21} : \frac{1012}{21}

= \frac{1012}{21}

\frac{d}{d x} (- x^{3} + 9 x^{2} - 24 x + 5)

\frac{\partial}{\partial x} (3 x^{2} - 2 x y - 4 y^{2})

\frac{d y}{d t} \cdot \frac{1}{4} + y = cos (t)

\int - \frac{1}{3} e^{- 3 t} d t

\frac{d}{d x} (3 x + e^{x})

What is the integral from 0 to 4 of 6x+1/((x+3)^2) ?
The integral from 0 to 4 of 6x+1/((x+3)^2) is 1012/21