What is the integral of 9(9x+4) ?

The integral of 9(9x+4) is 9((9x^2)/2+4x)+C

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integral of 9(9x+4)

\int 9 (9 x + 4) d x

9 (\frac{9 x ^{2}}{2} + 4 x) + C

Solution steps

\int 9 (9 x + 4) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int 9 x + 4 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

= 9 (\int 9 x d x + \int 4 d x)

\int 9 x d x = \frac{9 x ^{2}}{2}

\int 4 d x = 4 x

= 9 (\frac{9 x ^{2}}{2} + 4 x)

Add a constant to the solution

= 9 (\frac{9 x ^{2}}{2} + 4 x) + C

d er i v a t i v e f (x) = x^{2} - 2 cos (x) - 3

x y^{2} y^{^{'}} = y^{3} - x^{3}

\int \frac{( 1 )}{25 x ^{2} - 16} d x

\frac{d y}{d x} + 6 y = e^{x}

\int x^{3} + x^{\frac{1}{3}} d x