What is the integral of 6x+1/(x^2) ?

The integral of 6x+1/(x^2) is 3x^2-1/x+C

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integral of 6x+1/(x^2)

\int 6 x + \frac{1}{x ^{2}} d x

3 x^{2} - \frac{1}{x} + C

Solution steps

\int 6 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 6 x d x + \int \frac{1}{x ^{2}} d x

\int 6 x d x = 3 x^{2}

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

= 3 x^{2} - \frac{1}{x}

Add a constant to the solution

= 3 x^{2} - \frac{1}{x} + C

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

x \to 1 + lim (2 x - 3)

\int_{l n (2)}^{l n (4)} e^{- x} d x

\frac{d}{d x} (\frac{e ^{x}}{sin ( x )})

h (x) = lo g_{10} (2 x^{2} + 5 x + 2)