What is the integral of (sqrt(x)+1)^2 ?

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

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integral of (sqrt(x)+1)^2

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

\frac{x ^{2}}{2} + \frac{4}{3} x^{\frac{3}{2}} + x + C

Solution steps

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

Expand

(x + 1)^{2} : x + 2 x + 1

= \int x + 2 x + 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 d x + \int 2 x d x + \int 1 d x

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

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

\int 1 d x = x

= \frac{x ^{2}}{2} + \frac{4}{3} x^{\frac{3}{2}} + x

Add a constant to the solution

= \frac{x ^{2}}{2} + \frac{4}{3} x^{\frac{3}{2}} + x + C

x^{3} y^{2} = x y^{3} + 6

\int \frac{1}{x ^{2} x ^{2} - 15} d x

y = x^{2}, y = - x + 2

\frac{1}{( s - 1 ) ^{3}}

\int (9 x e^{x^{2}} + \frac{5}{x} - 5^{x} + 3 x^{2}) d x

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