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Exercise 7.8 · Q19

Q.Evaluate the definite integral: ∫026x+3x2+4 dx\int_{0}^{2} \frac{6x+3}{x^2+4} \, dx

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Split into a log part and an arctan part: the value is 3ln⁡2+3π83\ln2+\dfrac{3\pi}{8}.

Setting up

The denominator x2+4x^2+4 suggests an arctangent, but the numerator 6x+36x+3 is linear. The standard move is to break it into two pieces: one whose numerator is a multiple of ddx(x2+4)=2x\frac{d}{dx}(x^2+4)=2x (giving a logarithm) and one that is a constant over x2+4x^2+4 (giving an arctangent).

6x+3x2+4=6xx2+4+3x2+4.\frac{6x+3}{x^2+4}=\frac{6x}{x^2+4}+\frac{3}{x^2+4}.

1. The logarithm piece

With u=x2+4u=x^2+4, du=2x dxdu=2x\,dx, so 6x dx=3 du6x\,dx=3\,du; limits u:4→8u:4\to8:

∫026xx2+4 dx=∫483u du=3(ln⁡8−ln⁡4)=3ln⁡2.\int_0^2\frac{6x}{x^2+4}\,dx=\int_4^8\frac{3}{u}\,du=3(\ln8-\ln4)=3\ln2.

2. The arctangent piece …

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