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Question 129 of 129

Q.Evaluate: ∫2x+3x2+3x+7 dx\displaystyle\int \dfrac{2x+3}{x^2+3x+7}\, dx

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2026Subjective· 3mImportance★★★★★
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Since ddx(x2+3x+7)=2x+3\dfrac{d}{dx}(x^2+3x+7)=2x+3 exactly matches the numerator, the integral is simply the natural log of the denominator.

Let u=x2+3x+7u=x^2+3x+7. Then dudx=2x+3\dfrac{du}{dx}=2x+3, so du=(2x+3) dxdu=(2x+3)\,dx — exactly the numerator.

∫2x+3x2+3x+7 dx=∫duu=ln⁡∣u∣+C=ln⁡∣x2+3x+7∣+C\displaystyle\int\dfrac{2x+3}{x^2+3x+7}\,dx=\int\dfrac{du}{u}=\ln|u|+C=\ln|x^2+3x+7|+C

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