What is the integral of (10^x-1/(sqrt(x))) ?

The integral of (10^x-1/(sqrt(x))) is (10^x)/(ln(10))-2sqrt(x)+C

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integral of (10^x-1/(sqrt(x)))

\int (1 0^{x} - \frac{1}{x}) d x

\frac{1 0 ^{x}}{ln ( 10 )} - 2 x + C

Solution steps

\int 1 0^{x} - \frac{1}{x} 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 1 0^{x} d x - \int \frac{1}{x} d x

\int 1 0^{x} d x = \frac{1 0 ^{x}}{ln ( 10 )}

\int \frac{1}{x} d x = 2 x

= \frac{1 0 ^{x}}{ln ( 10 )} - 2 x

Add a constant to the solution

= \frac{1 0 ^{x}}{ln ( 10 )} - 2 x + C

\frac{d}{d x} (x^{2} + 6 x + 10)

\int 6 x^{4} - 6 x^{3} + 5 x^{2} + C d x

\frac{50}{1 - 3 s ^{2}}

\frac{\partial}{\partial x} (s^{8} t + s t^{- 6})

5 x y^{5} + 8 x y = 39, (3, 1)

What is the integral of (10^x-1/(sqrt(x))) ?
The integral of (10^x-1/(sqrt(x))) is (10^x)/(ln(10))-2sqrt(x)+C