What is the integral from 0 to x of (3+t^2)/(1-t^4) ?

The integral from 0 to x of (3+t^2)/(1-t^4) is arctan(x)+ln|x+1|-ln|x-1|

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integral from 0 to x of (3+t^2)/(1-t^4)

\int_{0}^{x} \frac{3 + t ^{2}}{1 - t ^{4}} d t

arctan (x) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣

Solution steps

\int_{0}^{x} \frac{3 + t ^{2}}{1 - t ^{4}} d t

Expand

\frac{3 + t ^{2}}{1 - t ^{4}} : \frac{3}{1 - t ^{4}} + \frac{t ^{2}}{1 - t ^{4}}

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} \frac{3}{1 - t ^{4}} d t + \int_{0}^{x} \frac{t ^{2}}{1 - t ^{4}} d t

\int_{0}^{x} \frac{3}{1 - t ^{4}} d t = 3 (\frac{1}{2} arctan (x) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣)

\int_{0}^{x} \frac{t ^{2}}{1 - t ^{4}} d t = - \frac{1}{2} arctan (x) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣

= 3 (\frac{1}{2} arctan (x) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣) - \frac{1}{2} arctan (x) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣

Simplify

3 (\frac{1}{2} arctan (x) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣) - \frac{1}{2} arctan (x) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣ : arctan (x) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣

= arctan (x) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣

\int_{1}^{2} \frac{x ^{2} - a}{x} d x

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

\int_{- 3}^{2} (9 - x^{2}) - (x + 3) d x

\int_{- 1}^{3} 3 x^{2} - 6 x - 9 d x

\int_{0}^{5} 1 + \frac{9}{4} x d x

What is the integral from 0 to x of (3+t^2)/(1-t^4) ?
The integral from 0 to x of (3+t^2)/(1-t^4) is arctan(x)+ln|x+1|-ln|x-1|