Evaluate the following integrals. Include absolute values only when needed. \int_{0}^{3}\frac{2x-1}{x+1}dx

prsategazd 2021-12-28 Answered
Evaluate the following integrals. Include absolute values only when needed.
032x1x+1dx
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Expert Answer

Thomas Lynn
Answered 2021-12-29 Author has 28 answers
Step 1:To determine
Evaluate:
032x1x+1dx
Step 2:Calculation
Consider the given integral:
032x1x+1dx
Let x+1=udx=du
Also, when x=0u=1 & when x=3u=4
So, the integral becomes:
142(u1)1udu
142u3udu
142uu3udu
1423udu
=2u3ln(u)14
=2(4)3ln(4)2(1)+3ln(1)
83ln(4)2+3(0)
63ln(22)
66ln(2)
Hence, 032x1x+1dx=66ln(2)
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zurilomk4
Answered 2021-12-30 Author has 35 answers
2x1x+1dx
Substitution u=x+1dudx=1dx=du:
=2u3udu
We use the distributive property:
=(23u)du
Lets
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karton
Answered 2022-01-04 Author has 368 answers

032x1x+1dx2x1x+1dx2t3tdt2tt3tdt23tdt2dt3tdt2t3ln(|t|)2(x+1)3ln(|x+1|)2x=23ln(|x+1|)2x+23ln(|x+1|)(2x+23ln(|x+1|))|032×3+23ln(|3+1|)(2×0+23ln(|0+1|))
Simplify
66ln(2)

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