"Evaluate the following. ∫(10y+11)dy4y2−4y+5" was answered by our community

necessaryh

Answered question

2021-02-25

Evaluate the following.

\int \frac{(10 y + 11) d y}{4 y^{2} - 4 y + 5}

Theodore Schwartz

Skilled2021-02-26Added 99 answers

The given integral is,

\int \frac{(10 y + 11) d y}{4 y^{2} - 4 y + 5}

First, complete the square in the denominator as follows:

\int \frac{(10 y + 11)}{4 y^{2} - 4 y + 5} d y = \int \frac{(10 y + 11)}{4 {(y - \frac{1}{2})}^{2} + 4} d y

Apply u-substitution,

u = y - \frac{1}{2}

\int \frac{(10 y + 11)}{4 {(y - \frac{1}{2})}^{2} + 4} d y = \int \frac{5 u + 8}{2 (u^{2} + 1)} d u

= \frac{1}{2} \int \frac{5 u + 8}{(u^{2} + 1)} d u

= \frac{1}{2} (\int \frac{5 u}{(u^{2} + 1)} d u + \int \frac{8}{(u^{2} + 1)} d u)

= \frac{1}{2} (\frac{5}{2} \ln | u^{2} + 1 | + 8 \tan^{- 1} (u))

Substitute back,

u = y - \frac{1}{2}

\frac{1}{2} (\frac{5}{2} \ln | u^{2} + 1 | + 8 \tan^{- 1} (u)) = \frac{1}{2} (\frac{5}{2} \ln | {(y - \frac{1}{2})}^{2} + 1 | + 8 \tan^{- 1} (y - \frac{1}{2}))

= \frac{1}{4} (5 \ln | y^{2} - y + \frac{5}{4} | + 16 \tan^{- 1} (y - \frac{1}{2})) + C

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