What is the integral of 10x^4+20x+8 ?

The integral of 10x^4+20x+8 is 2x^5+10x^2+8x+C

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integral of 10x^4+20x+8

\int 10 x^{4} + 20 x + 8 d x

2 x^{5} + 10 x^{2} + 8 x + C

Solution steps

\int 10 x^{4} + 20 x + 8 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 10 x^{4} d x + \int 20 x d x + \int 8 d x

\int 10 x^{4} d x = 2 x^{5}

\int 20 x d x = 10 x^{2}

\int 8 d x = 8 x

= 2 x^{5} + 10 x^{2} + 8 x

Add a constant to the solution

= 2 x^{5} + 10 x^{2} + 8 x + C

h \to 0 lim (\frac{( 1 + h ) ^{4} - 1}{h})

\int \frac{x ^{2}}{64 + x ^{2}} d x

\frac{d}{d x} (lo g_{e} (2 x))

\int - \frac{3}{x ^{2}} + 11 - 4^{x} d x

x \to - π + lim (csc (x))

What is the integral of 10x^4+20x+8 ?
The integral of 10x^4+20x+8 is 2x^5+10x^2+8x+C