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evaluate to 1.dp integral of 3 and 1 for 2/2x+5 dx
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This is a simple application of the natural logarithm. Give it a go.
i obviously have lol
Not a helpful answer. \(\int \dfrac{1}{x}\;dx = \ln|x| + C\) Utilize this result.
Try a u substitution - let u = 2x + 1 then \[\frac{du}{dx} = 2\]so\[\frac{ du }{ 2} = dx\] Now put them into the integral, and change the limits. \[\large \int\limits_{1}^{3} \frac{ 2 }{ 2x+5 } dx = \int\limits_{a}^{b} \frac{ 2 }{u } \frac{ du }{2 }\]
To change the limits, put the old limits (x=1 and x=3) into the equation for u: u = 2x+1 To find the limits for u.
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