What is the integral of (2-8x)/(1+4x^2) ?

The integral of (2-8x)/(1+4x^2) is arctan(2x)-ln|4x^2+1|+C

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integral of (2-8x)/(1+4x^2)

\int \frac{2 - 8 x}{1 + 4 x ^{2}} d x

arctan (2 x) - ln 4 x^{2} + 1 + C

Solution steps

\int \frac{2 - 8 x}{1 + 4 x ^{2}} d x

Apply Integration By Parts

: arctan (2 x) (- 4 x + 1) - \int - 4 arctan (2 x) d x

= arctan (2 x) (- 4 x + 1) - \int - 4 arctan (2 x) d x

\int - 4 arctan (2 x) d x = - 4 x arctan (2 x) + ln 4 x^{2} + 1

= arctan (2 x) (- 4 x + 1) - (- 4 x arctan (2 x) + ln 4 x^{2} + 1)

Simplify

arctan (2 x) (- 4 x + 1) - (- 4 x arctan (2 x) + ln 4 x^{2} + 1) : arctan (2 x) - ln 4 x^{2} + 1

= arctan (2 x) - ln 4 x^{2} + 1

Add a constant to the solution

= arctan (2 x) - ln 4 x^{2} + 1 + C

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What is the integral of (2-8x)/(1+4x^2) ?
The integral of (2-8x)/(1+4x^2) is arctan(2x)-ln|4x^2+1|+C