What is the integral of 400(1+(2x)/(24+x^2)) ?

The integral of 400(1+(2x)/(24+x^2)) is 400(x+ln|24+x^2|)+C

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integral of 400(1+(2x)/(24+x^2))

\int 400 (1 + \frac{2 x}{24 + x ^{2}}) d x

400 (x + ln 24 + x^{2}) + C

Solution steps

\int 400 (1 + \frac{2 x}{24 + x ^{2}}) d x

Take the constant out:

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

= 400 \cdot \int 1 + \frac{2 x}{24 + 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

= 400 (\int 1 d x + \int \frac{2 x}{24 + x ^{2}} d x)

\int 1 d x = x

\int \frac{2 x}{24 + x ^{2}} d x = ln 24 + x^{2}

= 400 (x + ln 24 + x^{2})

Add a constant to the solution

= 400 (x + ln 24 + x^{2}) + C

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What is the integral of 400(1+(2x)/(24+x^2)) ?
The integral of 400(1+(2x)/(24+x^2)) is 400(x+ln|24+x^2|)+C