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Wize AP Calculus (AB) Textbook > Integration & Accumulation of Change

Antiderivatives of 1x\frac{1}{x}

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Antiderivatives (Indefinite Integral) of 1x\color{blue}\frac{1}{x}

Let kk be a nonzero constant. Then we have the following indefinite integral rules:
∫1x dx=ln⁡∣x∣+C, (for x≠0)\boxed{\displaystyle\int\dfrac{1}{x} \ dx=\ln|x|+C,\ \text{(for}\ x\neq0)}

∫1kx+adx=1kln⁡∣kx+a∣+C,(for x≠0,a constant)\displaystyle \boxed{\int\frac{1}{kx+a}dx=\frac{1}{k}\ln|kx+a|+C }, \text{(for } x\ne0, a \text{ constant})

Watch Out!
Notice the absolute value bars inside the natural logarithm!

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Example: Indefinite integrals of 1x\color{blue}\frac{1}{x}

Evaluate the following indefinite integral

∫x2+x+2x2dx\displaystyle \int \frac{x^2+x+2}{x^2}dx

∫x2+x+2x2dx\displaystyle \int \frac{x^2+x+2}{x^2}dx

=∫(x2x2+xx2+2x2)dx=\displaystyle \int \bigg(\frac{x^2}{x^2}+\frac{x}{x^2}+\frac{2}{x^2}\bigg)dx

=∫(1+1x+2x−2)dx=\displaystyle \int(1+\frac{1}{x}+2x^{-2})dx

=x+ln⁡∣x∣+2x−1−1+C\displaystyle = x+\ln |x|+\frac{2x^{-1}}{-1}+C

=x+ln⁡∣x∣−2x+C\displaystyle =\boxed{ x+\ln |x|-\frac{2}{x}+C}

Evaluate the following indefinite integral

∫34−2x dx\displaystyle \int \frac{3}{4-2x} \ dx