Evaluate the following integrals. int_0^3frac{2x-1}{x+1}

aflacatn

aflacatn

Answered question

2021-02-12

Evaluate the following integrals.
032x1x+1

Answer & Explanation

Mayme

Mayme

Skilled2021-02-13Added 103 answers

Consider the given integral
032x1x+1
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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