What is the integral of 7-3(1+x^2)^{-1} ?

The integral of 7-3(1+x^2)^{-1} is 7x-3arctan(x)+C

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integral of 7-3(1+x^2)^{-1}

\int 7 - 3 (1 + x^{2})^{- 1} d x

7 x - 3 arctan (x) + C

Solution steps

\int 7 - 3 (1 + x^{2})^{- 1} 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 7 d x - \int 3 (1 + x^{2})^{- 1} d x

\int 7 d x = 7 x

\int 3 (1 + x^{2})^{- 1} d x = 3 arctan (x)

= 7 x - 3 arctan (x)

Add a constant to the solution

= 7 x - 3 arctan (x) + C

6 x^{^{''}} + 1960 x = 0, x (0) = - 3, x^{^{'}} (0) = 0

f (x) = (3)^{x}, at x = 8

x \to 2 lim (2^{c \cdot x})

\frac{d}{d x} (\frac{6 x}{2 x ^{\frac{1}{3}}})

y^{^{''}} + y = sin (x)

What is the integral of 7-3(1+x^2)^{-1} ?
The integral of 7-3(1+x^2)^{-1} is 7x-3arctan(x)+C