What is the integral of (x^{3/2}+4x+6) ?

The integral of (x^{3/2}+4x+6) is 2/5 x^{5/2}+2x^2+6x+C

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integral of (x^{3/2}+4x+6)

\int (x^{\frac{3}{2}} + 4 x + 6) d x

\frac{2}{5} x^{\frac{5}{2}} + 2 x^{2} + 6 x + C

Solution steps

\int x^{\frac{3}{2}} + 4 x + 6 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 x^{\frac{3}{2}} d x + \int 4 x d x + \int 6 d x

\int x^{\frac{3}{2}} d x = \frac{2}{5} x^{\frac{5}{2}}

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

\int 6 d x = 6 x

= \frac{2}{5} x^{\frac{5}{2}} + 2 x^{2} + 6 x

Add a constant to the solution

= \frac{2}{5} x^{\frac{5}{2}} + 2 x^{2} + 6 x + C

\int \frac{1}{4} e^{- \frac{x}{4}} d x

y = f (x), at x = 2

x \to - 2 - lim (\frac{- 3}{x ^{2} - 4})

n = 1 \sum \infty \frac{1}{n ^{3} + 2}

f (x) = x^{2} (1 - 7 x)

What is the integral of (x^{3/2}+4x+6) ?
The integral of (x^{3/2}+4x+6) is 2/5 x^{5/2}+2x^2+6x+C