What is the integral of-(2x^2)/5+(3x)/4-1/6 ?

The integral of-(2x^2)/5+(3x)/4-1/6 is -(2x^3}{15}+\frac{3x^2)/8-1/6 x+C

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integral of-(2x^2)/5+(3x)/4-1/6

\int - \frac{2 x ^{2}}{5} + \frac{3 x}{4} - \frac{1}{6} d x

- \frac{2 x ^{3}}{15} + \frac{3 x ^{2}}{8} - \frac{1}{6} x + C

Solution steps

\int - \frac{2 x ^{2}}{5} + \frac{3 x}{4} - \frac{1}{6} 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 \frac{2 x ^{2}}{5} d x + \int \frac{3 x}{4} d x - \int \frac{1}{6} d x

\int \frac{2 x ^{2}}{5} d x = \frac{2 x ^{3}}{15}

\int \frac{3 x}{4} d x = \frac{3 x ^{2}}{8}

\int \frac{1}{6} d x = \frac{1}{6} x

= - \frac{2 x ^{3}}{15} + \frac{3 x ^{2}}{8} - \frac{1}{6} x

Add a constant to the solution

= - \frac{2 x ^{3}}{15} + \frac{3 x ^{2}}{8} - \frac{1}{6} x + C

What is the integral of-(2x^2)/5+(3x)/4-1/6 ?
The integral of-(2x^2)/5+(3x)/4-1/6 is -(2x^3}{15}+\frac{3x^2)/8-1/6 x+C