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

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

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integral of (2x^3-3x+5)

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

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

Solution steps

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

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

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

\int 5 d x = 5 x

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

Add a constant to the solution

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

\int x cot (x^{2}) d x

y^{^{'}} + y = e^{- t}

n = 1 \sum \infty \frac{n - 5}{n} + 5

\int \frac{x ^{2}}{x ^{2} + 6} d x

\int (e^{- \frac{1}{x ^{2}} - 3 l n (x)} \cdot 1) d x

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