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Example · Example 9

Q.Evaluate ∫2x+3x2+4x+13 dx\displaystyle\int \frac{2x+3}{x^2+4x+13}\,dx.

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The denominator's derivative is 2x+42x+4. Write 2x+3=(2x+4)−12x+3=(2x+4)-1:

∫2x+3x2+4x+13 dx=∫2x+4x2+4x+13 dx−∫dxx2+4x+13.\int\frac{2x+3}{x^2+4x+13}\,dx = \int\frac{2x+4}{x^2+4x+13}\,dx - \int\frac{dx}{x^2+4x+13}.

The first integral is a pure substitution (u=x2+4x+13u=x^2+4x+13, du=(2x+4)dxdu=(2x+4)dx), giving

ln⁡∣x2+4x+13∣=ln⁡(x2+4x+13)\ln|x^2+4x+13|=\ln(x^2+4x+13) (always positive, discriminant 16−52<016-52<0). The second is Example 8's

result. So

∫2x+3x2+4x+13 dx=ln⁡(x2+4x+13)−13tan⁡−1 ⁣(x+23)+C.\int\frac{2x+3}{x^2+4x+13}\,dx = \ln(x^2+4x+13)-\frac13\tan^{-1}\!\left(\frac{x+2}{3}\right)+C.

Check: the derivative of the first term is 2x+4x2+4x+13\dfrac{2x+4}{x^2+4x+13}, and (from Example 8) the …

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