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

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

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integral of (9x+cos(x))

\int (9 x + cos (x)) d x

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

Solution steps

\int 9 x + cos (x) 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 9 x d x + \int cos (x) d x

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

\int cos (x) d x = sin (x)

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

Add a constant to the solution

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

x \to - 4 lim (x - 4)

\frac{d}{d x} (I n (x + x^{2} + 1))

3 x^{2} - 7 x^{3} + x^{4} - 5 - 3 x

(1.2) (0.8)

y^{^{'}} = \frac{( t ^{2} + 3 y ^{2} )}{2 t y}

What is the integral of (9x+cos(x)) ?
The integral of (9x+cos(x)) is (9x^2)/2+sin(x)+C