Two standard forms A handful of rational forms show up often enough in calculus that it's worth knowing their antiderivatives directly, rather than re-deriving them via partial fractions every time:
∫ d x x 2 − a 2 = 1 2 a ln ∣ x − a x + a ∣ + C \int \dfrac{dx}{x^2-a^2} = \dfrac{1}{2a}\ln\left|\dfrac{x-a}{x+a}\right| + C ∫ x 2 − a 2 d x = 2 a 1 ln x + a x − a + C ∫ d x a 2 − x 2 = 1 2 a ln ∣ a + x a − x ∣ + C \int \dfrac{dx}{a^2-x^2} = \dfrac{1}{2a}\ln\left|\dfrac{a+x}{a-x}\right| + C ∫ a 2 − x 2 d x = 2 a 1 ln a − x a + x + C Worked example Find ∫ x 2 1 − x 6 d x \displaystyle\int \dfrac{x^2}{1-x^6}\,dx ∫ 1 − x 6 x 2 d x
Solution: Substitute u = x 3 u = x^3 u = x 3 d u = 3 x 2 d x du = 3x^2\,dx d u = 3 x 2 d x
∫ x 2 1 − x 6 d x = 1 3 ∫ d u 1 − u 2 = 1 3 ⋅ 1 2 ln ∣ 1 + u 1 − u ∣ + C = 1 6 ln ∣ 1 + x 3 1 − x 3 ∣ + C \int \dfrac{x^2}{1-x^6}\,dx = \dfrac{1}{3}\int \dfrac{du}{1-u^2} = \dfrac{1}{3}\cdot\dfrac{1}{2}\ln\left|\dfrac{1+u}{1-u}\right| + C = \dfrac{1}{6}\ln\left|\dfrac{1+x^3}{1-x^3}\right| + C ∫ 1 − x 6 x 2 d x = 3 1 ∫ 1 − u 2 d u = 3 1 ⋅ 2 1 ln 1 − u 1 + u + C = 6 1 ln 1 − x 3 1 + x 3 + C