What is the integral of 7x^4-5x^3+4x^2+C ?

The integral of 7x^4-5x^3+4x^2+C is (7x^5)/5-(5x^4)/4+(4x^3)/3+Cx+C

Popular >

integral of 7x^4-5x^3+4x^2+C

\int 7 x^{4} - 5 x^{3} + 4 x^{2} + C d x

\frac{7 x ^{5}}{5} - \frac{5 x ^{4}}{4} + \frac{4 x ^{3}}{3} + C x + C

Solution steps

\int 7 x^{4} - 5 x^{3} + 4 x^{2} + C 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 7 x^{4} d x - \int 5 x^{3} d x + \int 4 x^{2} d x + \int C d x

\int 7 x^{4} d x = \frac{7 x ^{5}}{5}

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

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

\int C d x = C x

= \frac{7 x ^{5}}{5} - \frac{5 x ^{4}}{4} + \frac{4 x ^{3}}{3} + C x

Add a constant to the solution

= \frac{7 x ^{5}}{5} - \frac{5 x ^{4}}{4} + \frac{4 x ^{3}}{3} + C x + C

(x + 2) y^{^{'}} - y = \frac{3}{x + 2}

x \to \infty + lim (\frac{1}{x})

(\frac{t}{y}) d y + (1 + ln (y)) d t = 0

\int \frac{1}{6} sec (x) tan (x) d x

\int 5 tan (3 x) ln (sec (3 x)) d x

What is the integral of 7x^4-5x^3+4x^2+C ?
The integral of 7x^4-5x^3+4x^2+C is (7x^5)/5-(5x^4)/4+(4x^3)/3+Cx+C