Evaluate ∫1e2x+3x2+3x dx\displaystyle\int_1^e \frac{2x + 3}{x^2 + 3x} \, dx∫1ex2+3x2x+3dx
2e+3−ln(e+3)2e + 3 - \ln(e + 3)2e+3−ln(e+3)
e2+3e−4e^2 + 3e - 4e2+3e−4
e2+3e4\frac{e^2 + 3e}{4}4e2+3e
ln(e2+3e)−ln4\ln(e^2 + 3e) - \ln 4ln(e2+3e)−ln4
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AI Tutor 5:04 AM
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