What is the integral of (5x^4+x/7) ?

The integral of (5x^4+x/7) is x^5+(x^2)/(14)+C

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integral of (5x^4+x/7)

\int (5 x^{4} + \frac{x}{7}) d x

x^{5} + \frac{x ^{2}}{14} + C

Solution steps

\int 5 x^{4} + \frac{x}{7} 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 5 x^{4} d x + \int \frac{x}{7} d x

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

\int \frac{x}{7} d x = \frac{x ^{2}}{14}

= x^{5} + \frac{x ^{2}}{14}

Add a constant to the solution

= x^{5} + \frac{x ^{2}}{14} + C

\frac{d y}{d x} + y + 2 cos (x) = 0

\frac{d}{d x} (\frac{( x ^{2} + 3 )}{x})

\frac{s}{s ^{2} + 9}

\frac{d}{d x} (\frac{2 x - 9}{x})

\int 3 x^{3} e^{- x^{2}} d x

What is the integral of (5x^4+x/7) ?
The integral of (5x^4+x/7) is x^5+(x^2)/(14)+C