What is the integral of (x^{3/2}+1)^2 ?

The integral of (x^{3/2}+1)^2 is (x^4)/4+4/5 x^{5/2}+x+C

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integral of (x^{3/2}+1)^2

\int (x^{\frac{3}{2}} + 1)^{2} d x

\frac{x ^{4}}{4} + \frac{4}{5} x^{\frac{5}{2}} + x + C

Solution steps

\int (x^{\frac{3}{2}} + 1)^{2} d x

Expand

(x^{\frac{3}{2}} + 1)^{2} : x^{3} + 2 x^{\frac{3}{2}} + 1

= \int x^{3} + 2 x^{\frac{3}{2}} + 1 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^{3} d x + \int 2 x^{\frac{3}{2}} d x + \int 1 d x

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

\int 2 x^{\frac{3}{2}} d x = \frac{4}{5} x^{\frac{5}{2}}

\int 1 d x = x

= \frac{x ^{4}}{4} + \frac{4}{5} x^{\frac{5}{2}} + x

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{4}{5} x^{\frac{5}{2}} + x + C

f (x) = sin (sin (sin (x)))

\frac{d y}{d x}, e^{x y} + x^{2} - y^{2} = 12

\frac{\partial}{\partial θ} (sin (θ))

\int e^{- x} + 4^{x} d x

\frac{d}{d x} (a \cdot ln (x))

What is the integral of (x^{3/2}+1)^2 ?
The integral of (x^{3/2}+1)^2 is (x^4)/4+4/5 x^{5/2}+x+C