What is the integral from 0 to x^2 of (4t-1)/(t^2+1) ?

The integral from 0 to x^2 of (4t-1)/(t^2+1) is 2ln|x^4+1|-arctan(x^2)

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integral from 0 to x^2 of (4t-1)/(t^2+1)

\int_{0}^{x^{2}} \frac{4 t - 1}{t ^{2} + 1} d t

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

Solution steps

\int_{0}^{x^{2}} \frac{4 t - 1}{t ^{2} + 1} d t

Expand

\frac{4 t - 1}{t ^{2} + 1} : \frac{4 t}{t ^{2} + 1} - \frac{1}{t ^{2} + 1}

Apply the Sum Rule:

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

= \int_{0}^{x^{2}} \frac{4 t}{t ^{2} + 1} d t - \int_{0}^{x^{2}} \frac{1}{t ^{2} + 1} d t

\int_{0}^{x^{2}} \frac{4 t}{t ^{2} + 1} d t = 2 ln x^{4} + 1

\int_{0}^{x^{2}} \frac{1}{t ^{2} + 1} d t = arctan (x^{2})

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

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What is the integral from 0 to x^2 of (4t-1)/(t^2+1) ?
The integral from 0 to x^2 of (4t-1)/(t^2+1) is 2ln|x^4+1|-arctan(x^2)