What is the integral of x^{100}+x^{99} ?

The integral of x^{100}+x^{99} is (x^{101})/(101)+(x^{100})/(100)+C

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integral of x^{100}+x^{99}

\int x^{100} + x^{99} d x

\frac{x ^{101}}{101} + \frac{x ^{100}}{100} + C

Solution steps

\int x^{100} + x^{99} 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 x^{100} d x + \int x^{99} d x

\int x^{100} d x = \frac{x ^{101}}{101}

\int x^{99} d x = \frac{x ^{100}}{100}

= \frac{x ^{101}}{101} + \frac{x ^{100}}{100}

Add a constant to the solution

= \frac{x ^{101}}{101} + \frac{x ^{100}}{100} + C

(x^{2} - y^{2}) d x + 2 x y d y = 0

\int_{0}^{1} t^{3} (3 + t^{4})^{3} d t

f (x) = 12 - x

t \to 0 lim (\frac{sin ( 3 t )}{t})

y = 3 x + 6 cos (x), (\frac{π}{3}, π + 3)

What is the integral of x^{100}+x^{99} ?
The integral of x^{100}+x^{99} is (x^{101})/(101)+(x^{100})/(100)+C