Expert Answer to "Evaluate the following integrals. ∫032x−1x+1"

aflacatn

Answered question

2021-02-12

Evaluate the following integrals.

\int_{0}^{3} \frac{2 x - 1}{x + 1}

Mayme

Skilled2021-02-13Added 103 answers

Consider the given integral

\int_{0}^{3} \frac{2 x - 1}{x + 1}

Let x + 1 = u \Rightarrow d x = d u

Also, when x = 0 \Rightarrow u = 1 when x = 3 \Rightarrow u = 4

So, the integral becomes:

\int_{1}^{4} \frac{2 (u - 1) - 1}{u} d u

\Rightarrow \int_{1}^{4} \frac{2 u - 3}{u} d u

\Rightarrow \int_{1}^{4} \frac{2 u}{u} - \frac{3}{u} d u

\Rightarrow \int_{1}^{4} 2 - \frac{3}{u} d u

= 2 u - 3 \ln (u) |_{1}^{4}

= 2 (4) - 3 \ln (4) - 2 (1) + 3 \ln (1)

\Rightarrow 8 - 3 \ln (4) - 2 + 3 (0)

\Rightarrow 6 - 3 \ln (2^{2})

\Rightarrow 6 - 6 \ln (2)

Hence, \int_{0}^{3} \frac{2 x - 1}{x + 1} d x = 6 - 6 \ln (2)

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