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

The integral of (1/5 x^3+3x^2-3/4 x+4) is (x^4}{20}+x^3-\frac{3x^2)/8+4x+C

Popular >

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

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

\frac{x ^{4}}{20} + x^{3} - \frac{3 x ^{2}}{8} + 4 x + C

Solution steps

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

\int \frac{1}{5} x^{3} d x = \frac{x ^{4}}{20}

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

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

\int 4 d x = 4 x

= \frac{x ^{4}}{20} + x^{3} - \frac{3 x ^{2}}{8} + 4 x

Add a constant to the solution

= \frac{x ^{4}}{20} + x^{3} - \frac{3 x ^{2}}{8} + 4 x + C

\int \frac{1 - 2 x}{1 - x ^{2}} d x

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

\int \frac{ln ( ln ( 6 x ) )}{x ln ( 6 x )} d x

\int e^{x} + 8 sin (x) d x

\int \frac{- x - 28}{( x - 2 ) ( x ^{2} - 2 x + 10 )} d x

What is the integral of (1/5 x^3+3x^2-3/4 x+4) ?
The integral of (1/5 x^3+3x^2-3/4 x+4) is (x^4}{20}+x^3-\frac{3x^2)/8+4x+C