Solved: "Evaluate the following integral: ∫(y−3)y2−6y+1"

permaneceerc

Answered question

2021-05-11

Evaluate the following integral:

\int \frac{(y - 3)}{y^{2} - 6 y + 1}

Layton

Skilled2021-05-12Added 89 answers

Substitute

y^{2} - 6 y + 1 = t

differentiate it with respect to y,

2 y - 6 = \frac{d t}{d y}

2 (y - 3) = \frac{d t}{d y}

(y - 3) d y = \frac{d t}{2}

Hence the integral becomes:

I = \int \frac{y - 3}{y^{2} - 6 y + 1} d y

= \frac{1}{2} \int \frac{d t}{t}

= \frac{1}{2} \log | t | + c

= \frac{1}{2} \log | y^{2} - 6 y + 1 | + c

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