Solved: "Evaluate the integral. ∫12(x+1)2xdx"

Kye

Answered question

2021-05-21

Evaluate the integral.

\int_{1}^{2} \frac{(x + 1)^{2}}{x} d x

cyhuddwyr9

Skilled2021-05-22Added 90 answers

Step 1
We need to evaluate the following integral:

\int_{1}^{2} \frac{(x + 1)^{2}}{x} d x

Expand the expression in the numerator:

\int_{1}^{2} \frac{(x^{2} + 2 x + 1)}{x} d x

Step 2
The integral in Step 1 can be rewritten as:

\int_{1}^{2} (\frac{x^{2}}{x} + \frac{2 x}{x} + \frac{1}{x}) d x

Using sum rule for integrals, we get:

\int_{1}^{2} \frac{x^{2}}{x} d x + \int_{1}^{2} \frac{2 x}{x} d x + \int_{1}^{2} \frac{1}{x} d x

\Rightarrow \int_{1}^{2} x d x + \int_{1}^{2} 2 d x + \int_{1}^{2} \frac{1}{x} d x

Solve each integral :

\Rightarrow [\frac{x^{2}}{2}]_{1}^{2} + 2 [x]_{1}^{2} + [\ln (x)]_{1}^{2}

\Rightarrow [\frac{4}{2} - \frac{1}{2}] + 2 [2 - 1] + [\ln (2) - \ln (1)]

\Rightarrow \frac{3}{2} + 2 + \ln (2)

\Rightarrow \frac{7}{2} + \ln (2)

Step 3
Answer:

\frac{7}{2} + \ln (2)

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