What is the integral of (ax^4+bx^3+3c) ?

The integral of (ax^4+bx^3+3c) is (ax^5)/5+(bx^4)/4+3cx+C

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integral of (ax^4+bx^3+3c)

\int (a x^{4} + b x^{3} + 3 c) d x

\frac{a x ^{5}}{5} + \frac{b x ^{4}}{4} + 3 c x + C

Solution steps

\int a x^{4} + b x^{3} + 3 c 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 a x^{4} d x + \int b x^{3} d x + \int 3 c d x

\int a x^{4} d x = \frac{a x ^{5}}{5}

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

\int 3 c d x = 3 c x

= \frac{a x ^{5}}{5} + \frac{b x ^{4}}{4} + 3 c x

Add a constant to the solution

= \frac{a x ^{5}}{5} + \frac{b x ^{4}}{4} + 3 c x + C

\frac{\partial}{\partial x} (\frac{( 2 π \cdot l ^{2} )}{x})

\int_{0}^{16} x d x

\int_{0}^{4} 4 y^{2} - \frac{y ^{4}}{4} d y

x \to a lim (f (x) + g (x))

f (t) = cos (2 t)

What is the integral of (ax^4+bx^3+3c) ?
The integral of (ax^4+bx^3+3c) is (ax^5)/5+(bx^4)/4+3cx+C