What is the integral of (1+sqrt(x))^5 ?

The integral of (1+sqrt(x))^5 is x+10/3 x^{3/2}+5x^2+4x^{5/2}+(5x^3)/3+2/7 x^{7/2}+C

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integral of (1+sqrt(x))^5

\int (1 + x)^{5} d x

x + \frac{10}{3} x^{\frac{3}{2}} + 5 x^{2} + 4 x^{\frac{5}{2}} + \frac{5 x ^{3}}{3} + \frac{2}{7} x^{\frac{7}{2}} + C

Solution steps

\int (1 + x)^{5} d x

Expand

(1 + x)^{5} : 1 + 5 x + 10 x + 10 x^{\frac{3}{2}} + 5 x^{2} + x^{\frac{5}{2}}

= \int 1 + 5 x + 10 x + 10 x^{\frac{3}{2}} + 5 x^{2} + x^{\frac{5}{2}} 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 1 d x + \int 5 x d x + \int 10 x d x + \int 10 x^{\frac{3}{2}} d x + \int 5 x^{2} d x + \int x^{\frac{5}{2}} d x

\int 1 d x = x

\int 5 x d x = \frac{10}{3} x^{\frac{3}{2}}

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

\int 10 x^{\frac{3}{2}} d x = 4 x^{\frac{5}{2}}

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

\int x^{\frac{5}{2}} d x = \frac{2}{7} x^{\frac{7}{2}}

= x + \frac{10}{3} x^{\frac{3}{2}} + 5 x^{2} + 4 x^{\frac{5}{2}} + \frac{5 x ^{3}}{3} + \frac{2}{7} x^{\frac{7}{2}}

Add a constant to the solution

= x + \frac{10}{3} x^{\frac{3}{2}} + 5 x^{2} + 4 x^{\frac{5}{2}} + \frac{5 x ^{3}}{3} + \frac{2}{7} x^{\frac{7}{2}} + C

x \to \infty lim (\frac{5 x}{5 x + 3 x ^{\frac{1}{x}}})

f (x) = 2 x - 3, 4, 0

f (x) = \frac{2}{x}, at x = 5

e^{3 x}

\int \frac{x + 5}{x + 6} d x

What is the integral of (1+sqrt(x))^5 ?
The integral of (1+sqrt(x))^5 is x+10/3 x^{3/2}+5x^2+4x^{5/2}+(5x^3)/3+2/7 x^{7/2}+C