What is the integral of cos(x)+1/3 sin(x)-1/(x^2) ?

The integral of cos(x)+1/3 sin(x)-1/(x^2) is sin(x)-1/3 cos(x)+1/x+C

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integral of cos(x)+1/3 sin(x)-1/(x^2)

\int cos (x) + \frac{1}{3} sin (x) - \frac{1}{x ^{2}} d x

sin (x) - \frac{1}{3} cos (x) + \frac{1}{x} + C

Solution steps

\int cos (x) + \frac{1}{3} sin (x) - \frac{1}{x ^{2}} 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 cos (x) d x + \int \frac{1}{3} sin (x) d x - \int \frac{1}{x ^{2}} d x

\int cos (x) d x = sin (x)

\int \frac{1}{3} sin (x) d x = - \frac{1}{3} cos (x)

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

= sin (x) - \frac{1}{3} cos (x) - (- \frac{1}{x})

Simplify

= sin (x) - \frac{1}{3} cos (x) + \frac{1}{x}

Add a constant to the solution

= sin (x) - \frac{1}{3} cos (x) + \frac{1}{x} + C

\frac{d}{d x} ((2^{x + 1} - 3. 5^{x})^{3})

\int 5 x + sin (x) d x

d er i v a t i v e \frac{1}{c}

t an g e n t f (x) = tan (x) + x, at x = \frac{π}{4}

x \to - 2 lim (- 6)

What is the integral of cos(x)+1/3 sin(x)-1/(x^2) ?
The integral of cos(x)+1/3 sin(x)-1/(x^2) is sin(x)-1/3 cos(x)+1/x+C