Step by step solution to "Evaluate the integrals. ∫23x2+2x−2x3+3x2−6xdx"

allhvasstH

Answered question

2021-10-27

Evaluate the integrals.

\int_{2}^{3} \frac{x^{2} + 2 x - 2}{x^{3} + 3 x^{2} - 6 x} d x

Nathaniel Kramer

Skilled2021-10-28Added 78 answers

The given integral is:

\int_{2}^{3} \frac{x^{2} + 2 x - 2}{x^{3} + 3 x^{2} - 6 x} d x

we have to evaluate the given integral.
Let the given integral be I.
therefore

I = \int_{2}^{3} \frac{x^{2} + 2 x - 2}{x^{3} + 3 x^{2} - 6 x} d x

let

x^{3} + 3 x^{2} - 6 x = t

therefore,

d (x^{3} + 3 x^{2} - 6 x) = d t

(3 x^{2} + 6 x - 6) d x = d t

3 (x^{2} + 2 x - 2) d x = \frac{d t}{3}

when x=2,

x^{3} + 3 x^{2} - 6 x = t

2^{3} + 3 {(2)}^{2} - 6 (2) = t

8 + 12 - 12 = t

8 = t

Therefore now the lower limit of the integral is 8
when x=3,

x^{3} + 3 x^{2} - 6 x = t

3^{3} + 3 {(3)}^{2} - 6 (3) = t

27 + 27 - 18 = t

27 + 9 = t

36 = t

therefore now the upper limit of the integral is 36.
now substitute these values in the integral I.
therefore,

I = \int_{2}^{3} \frac{x^{2} + 2 x - 2}{x^{3} + 3 x^{2} - 6 x} d x

= \int_{8}^{36} \frac{d t}{3 t}

= \frac{1}{3} \int_{8}^{36} \frac{d t}{t}

= \frac{1}{3} {[\ln t]}_{8}^{36}

= \frac{1}{3} {[\ln t]}_{8}^{36}

= \frac{1}{3} [\ln 36 - \ln 8]

= \frac{1}{3} [\ln \frac{36}{8}]

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