What is the integral of (sqrt(x)+2)^4 ?

The integral of (sqrt(x)+2)^4 is (x^3)/3+16/5 x^{5/2}+12x^2+64/3 x^{3/2}+16x+C

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integral of (sqrt(x)+2)^4

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

\frac{x ^{3}}{3} + \frac{16}{5} x^{\frac{5}{2}} + 12 x^{2} + \frac{64}{3} x^{\frac{3}{2}} + 16 x + C

Solution steps

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

Expand

(x + 2)^{4} : x^{2} + 8 x^{\frac{3}{2}} + 24 x + 32 x + 16

= \int x^{2} + 8 x^{\frac{3}{2}} + 24 x + 32 x + 16 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^{2} d x + \int 8 x^{\frac{3}{2}} d x + \int 24 x d x + \int 32 x d x + \int 16 d x

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

\int 8 x^{\frac{3}{2}} d x = \frac{16}{5} x^{\frac{5}{2}}

\int 24 x d x = 12 x^{2}

\int 32 x d x = \frac{64}{3} x^{\frac{3}{2}}

\int 16 d x = 16 x

= \frac{x ^{3}}{3} + \frac{16}{5} x^{\frac{5}{2}} + 12 x^{2} + \frac{64}{3} x^{\frac{3}{2}} + 16 x

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{16}{5} x^{\frac{5}{2}} + 12 x^{2} + \frac{64}{3} x^{\frac{3}{2}} + 16 x + C

What is the integral of (sqrt(x)+2)^4 ?
The integral of (sqrt(x)+2)^4 is (x^3)/3+16/5 x^{5/2}+12x^2+64/3 x^{3/2}+16x+C