What is the integral of 5000(1-(100)/((x+10)^2)) ?

The integral of 5000(1-(100)/((x+10)^2)) is 5000(x+(100)/(x+10))+C

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integral of 5000(1-(100)/((x+10)^2))

\int 5000 (1 - \frac{100}{( x + 10 ) ^{2}}) d x

5000 (x + \frac{100}{x + 10}) + C

Solution steps

\int 5000 (1 - \frac{100}{( x + 10 ) ^{2}}) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5000 \cdot \int 1 - \frac{100}{( x + 10 ) ^{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

= 5000 (\int 1 d x - \int \frac{100}{( x + 10 ) ^{2}} d x)

\int 1 d x = x

\int \frac{100}{( x + 10 ) ^{2}} d x = - \frac{100}{x + 10}

= 5000 (x - (- \frac{100}{x + 10}))

Simplify

= 5000 (x + \frac{100}{x + 10})

Add a constant to the solution

= 5000 (x + \frac{100}{x + 10}) + C

\frac{\partial}{\partial y} (\frac{9 x ^{2}}{y ^{\frac{1}{3}}})

\frac{d}{d x} (x^{4} - 2 x^{2} + 2)

f (x) = \frac{x ^{2}}{1 + 8 x}

- 5 y

3 (x^{3} - x), 0

What is the integral of 5000(1-(100)/((x+10)^2)) ?
The integral of 5000(1-(100)/((x+10)^2)) is 5000(x+(100)/(x+10))+C