What is the integral of (-3)/(x^2)+11-4^x ?

The integral of (-3)/(x^2)+11-4^x is 3/x+11x-(2^{2x-1})/(ln(2))+C

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integral of (-3)/(x^2)+11-4^x

\int \frac{- 3}{x ^{2}} + 11 - 4^{x} d x

\frac{3}{x} + 11 x - \frac{2 ^{2 x - 1}}{ln ( 2 )} + C

Solution steps

\int \frac{- 3}{x ^{2}} + 11 - 4^{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 \frac{- 3}{x ^{2}} d x + \int 11 d x - \int 4^{x} d x

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

\int 11 d x = 11 x

\int 4^{x} d x = \frac{2 ^{2 x - 1}}{ln ( 2 )}

= \frac{3}{x} + 11 x - \frac{2 ^{2 x - 1}}{ln ( 2 )}

Add a constant to the solution

= \frac{3}{x} + 11 x - \frac{2 ^{2 x - 1}}{ln ( 2 )} + C

sin^{5} (x^{3})

x \to 5 lim (∣ x - 5 ∣)

y^{^{'}} - \frac{y}{2} = e^{- t}

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

(x + 1) \frac{d y}{d x} = x + 6

What is the integral of (-3)/(x^2)+11-4^x ?
The integral of (-3)/(x^2)+11-4^x is 3/x+11x-(2^{2x-1})/(ln(2))+C