What is the integral of (10x^3-4x^2+2x-1) ?

The integral of (10x^3-4x^2+2x-1) is (5x^4)/2-(4x^3)/3+x^2-x+C

integral of (10x^3-4x^2+2x-1)

\int (10 x^{3} - 4 x^{2} + 2 x - 1) d x

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

Solution steps

\int 10 x^{3} - 4 x^{2} + 2 x - 1 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 10 x^{3} d x - \int 4 x^{2} d x + \int 2 x d x - \int 1 d x

\int 10 x^{3} d x = \frac{5 x ^{4}}{2}

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\int (sin (t^{2})) d t

\int t^{2} cos (nπ t) d t

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

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

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

What is the integral of (10x^3-4x^2+2x-1) ?
The integral of (10x^3-4x^2+2x-1) is (5x^4)/2-(4x^3)/3+x^2-x+C