What is the integral from 3x to 2 of (t^2-1)/(t^2+1) ?

The integral from 3x to 2 of (t^2-1)/(t^2+1) is -2(arctan(2)-arctan(3x))+2-3x

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integral from 3x to 2 of (t^2-1)/(t^2+1)

\int_{3 x}^{2} \frac{t ^{2} - 1}{t ^{2} + 1} d t

- 2 (arctan (2) - arctan (3 x)) + 2 - 3 x

Solution steps

\int_{3 x}^{2} \frac{t ^{2} - 1}{t ^{2} + 1} d t

\frac{t ^{2} - 1}{t ^{2} + 1} = - \frac{2}{t ^{2} + 1} + 1

= \int_{3 x}^{2} - \frac{2}{t ^{2} + 1} + 1 d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{3 x}^{2} \frac{2}{t ^{2} + 1} d t + \int_{3 x}^{2} 1 d t

\int_{3 x}^{2} \frac{2}{t ^{2} + 1} d t = 2 (arctan (2) - arctan (3 x))

\int_{3 x}^{2} 1 d t = 2 - 3 x

= - 2 (arctan (2) - arctan (3 x)) + 2 - 3 x

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What is the integral from 3x to 2 of (t^2-1)/(t^2+1) ?
The integral from 3x to 2 of (t^2-1)/(t^2+1) is -2(arctan(2)-arctan(3x))+2-3x