What is the integral of 4+10(x+5)^{-2} ?

The integral of 4+10(x+5)^{-2} is 4x-(10)/(x+5)+C

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integral of 4+10(x+5)^{-2}

\int 4 + 10 (x + 5)^{- 2} d x

4 x - \frac{10}{x + 5} + C

Solution steps

\int 4 + 10 (x + 5)^{- 2} d x

10 (x + 5)^{- 2} = \frac{10}{( x + 5 ) ^{2}}

= \int 4 + \frac{10}{( x + 5 ) ^{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 4 d x + \int \frac{10}{( x + 5 ) ^{2}} d x

\int 4 d x = 4 x

\int \frac{10}{( x + 5 ) ^{2}} d x = - \frac{10}{x + 5}

= 4 x - \frac{10}{x + 5}

Add a constant to the solution

= 4 x - \frac{10}{x + 5} + C

What is the integral of 4+10(x+5)^{-2} ?
The integral of 4+10(x+5)^{-2} is 4x-(10)/(x+5)+C