What is the integral of (3x^4+x^5) ?

The integral of (3x^4+x^5) is (3x^5)/5+(x^6)/6+C

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integral of (3x^4+x^5)

\int (3 x^{4} + x^{5}) d x

\frac{3 x ^{5}}{5} + \frac{x ^{6}}{6} + C

Solution steps

\int 3 x^{4} + x^{5} 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 3 x^{4} d x + \int x^{5} d x

\int 3 x^{4} d x = \frac{3 x ^{5}}{5}

\int x^{5} d x = \frac{x ^{6}}{6}

= \frac{3 x ^{5}}{5} + \frac{x ^{6}}{6}

Add a constant to the solution

= \frac{3 x ^{5}}{5} + \frac{x ^{6}}{6} + C

(1 + y^{2}) d x + x y d y = 0

x = \frac{7}{y}, x = \frac{y}{9}, x = 9 y

x \to 0 lim (2 x cot (x))

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

\frac{d y}{d x} = y^{2} (1 + x^{2})

What is the integral of (3x^4+x^5) ?
The integral of (3x^4+x^5) is (3x^5)/5+(x^6)/6+C