What is the integral of 4x^3+9sin(x) ?

The integral of 4x^3+9sin(x) is x^4-9cos(x)+C

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integral of 4x^3+9sin(x)

\int 4 x^{3} + 9 sin (x) d x

x^{4} - 9 cos (x) + C

Solution steps

\int 4 x^{3} + 9 sin (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 4 x^{3} d x + \int 9 sin (x) d x

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

\int 9 sin (x) d x = - 9 cos (x)

= x^{4} - 9 cos (x)

Add a constant to the solution

= x^{4} - 9 cos (x) + C

x \to 3 lim (x + 1)

\int sech^{2} (x) d x

n = 1 \sum \infty n! (x + 1)^{n}

n = 1 \sum \infty (- 1)^{n} \cdot \frac{1}{n}

x \to 0 - lim (\frac{( x - 1 )}{x ^{2}})

What is the integral of 4x^3+9sin(x) ?
The integral of 4x^3+9sin(x) is x^4-9cos(x)+C