What is the integral of x^{1/4}+9 ?

The integral of x^{1/4}+9 is 4/5 x^{5/4}+9x+C

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integral of x^{1/4}+9

\int x^{\frac{1}{4}} + 9 d x

\frac{4}{5} x^{\frac{5}{4}} + 9 x + C

Solution steps

\int x^{\frac{1}{4}} + 9 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^{\frac{1}{4}} d x + \int 9 d x

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

\int 9 d x = 9 x

= \frac{4}{5} x^{\frac{5}{4}} + 9 x

Add a constant to the solution

= \frac{4}{5} x^{\frac{5}{4}} + 9 x + C

\int sec (x) (9 sec (x) + 7 tan (x)) d x

\int \frac{x + 1}{1 + x} d x

\int (tan (x))^{2} (sec (x))^{2} d x

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

\int - 0.1 x d x

What is the integral of x^{1/4}+9 ?
The integral of x^{1/4}+9 is 4/5 x^{5/4}+9x+C