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

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

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

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

(x + 1) (\frac{2 x ^{3}}{3} + \frac{2}{3} x^{\frac{3}{2}}) - \frac{1}{3} (\frac{x ^{4}}{2} + \frac{4}{5} x^{\frac{5}{2}}) + C

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{1}{3} (2 x^{3} + 2 x^{\frac{3}{2}}) d x = \frac{1}{3} (\frac{x ^{4}}{2} + \frac{4}{5} x^{\frac{5}{2}})

= (x + 1) (\frac{2 x ^{3}}{3} + \frac{2}{3} x^{\frac{3}{2}}) - \frac{1}{3} (\frac{x ^{4}}{2} + \frac{4}{5} x^{\frac{5}{2}})

Add a constant to the solution

= (x + 1) (\frac{2 x ^{3}}{3} + \frac{2}{3} x^{\frac{3}{2}}) - \frac{1}{3} (\frac{x ^{4}}{2} + \frac{4}{5} x^{\frac{5}{2}}) + C

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} = 0

x \to - 3 lim (\frac{3 - ∣ x ∣}{3 + x})

y^{^{''}} + 6 y^{^{'}} + 34 y = 25 e^{- 3 x}

\frac{d}{d x} (4 x - 5 x^{\frac{5}{6}})

f (x) = x cos (x) + 2 tan (x)

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