What is the integral of 2x^2-3x+5 ?

The integral of 2x^2-3x+5 is (2x^3)/3-(3x^2)/2+5x+C

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integral of 2x^2-3x+5

\int 2 x^{2} - 3 x + 5 d x

\frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 5 x + C

Solution steps

\int 2 x^{2} - 3 x + 5 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 2 x^{2} d x - \int 3 x d x + \int 5 d x

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

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

\int 5 d x = 5 x

= \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 5 x

Add a constant to the solution

= \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 5 x + C

\frac{d y}{d t} + \frac{23 y}{t} = \frac{1}{t}

1390 - 34 x, 88 x, [11.3934400000000 \dots, 12.3934400000000 \dots]

f (x) = n! (3 - x)^{- 1}

\frac{\partial}{\partial x} (x y^{2} e^{y})

y^{^{''}} - 4 y^{^{'}} + 8 = 0

What is the integral of 2x^2-3x+5 ?
The integral of 2x^2-3x+5 is (2x^3)/3-(3x^2)/2+5x+C