Question

Evaluate the following. intfrac{(10y+11)dy}{4y^2-4y+5}

Trigonometry
ANSWERED
asked 2021-02-25
Evaluate the following.
\(\displaystyle\int{\frac{{{\left({10}{y}+{11}\right)}{\left.{d}{y}\right.}}}{{{4}{y}^{{2}}-{4}{y}+{5}}}}\)

Answers (1)

2021-02-26
The given integral is, \(\displaystyle\int{\frac{{{\left({10}{y}+{11}\right)}{\left.{d}{y}\right.}}}{{{4}{y}^{{2}}-{4}{y}+{5}}}}\)
First, complete the square in the denominator as follows:
\(\displaystyle\int{\frac{{{\left({10}{y}+{11}\right)}}}{{{4}{y}^{{2}}-{4}{y}+{5}}}}{\left.{d}{y}\right.}=\int{\frac{{{\left({10}{y}+{11}\right)}}}{{{4}{\left({y}-{\frac{{{1}}}{{{2}}}}\right)}^{{2}}+{4}}}}{\left.{d}{y}\right.}\)
Apply u-substitution, \(\displaystyle{u}={y}-{\frac{{{1}}}{{{2}}}}\)
\(\displaystyle\int{\frac{{{\left({10}{y}+{11}\right)}}}{{{4}{\left({y}-{\frac{{{1}}}{{{2}}}}\right)}^{{2}}+{4}}}}{\left.{d}{y}\right.}=\int{\frac{{{5}{u}+{8}}}{{{2}{\left({u}^{{2}}+{1}\right)}}}}{d}{u}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}\int{\frac{{{5}{u}+{8}}}{{{\left({u}^{{2}}+{1}\right)}}}}{d}{u}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\left(\int{\frac{{{5}{u}}}{{{\left({u}^{{2}}+{1}\right)}}}}{d}{u}+\int{\frac{{{8}}}{{{\left({u}^{{2}}+{1}\right)}}}}{d}{u}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\left({\frac{{{5}}}{{{2}}}}{\ln}{\left|{u}^{{2}}+{1}\right|}+{8}{{\tan}^{{-{1}}}{\left({u}\right)}}\right)}\)
Substitute back, \(\displaystyle{u}={y}-{\frac{{{1}}}{{{2}}}}\)
\(\displaystyle{\frac{{{1}}}{{{2}}}}{\left({\frac{{{5}}}{{{2}}}}{\ln}{\left|{u}^{{2}}+{1}\right|}+{8}{{\tan}^{{-{1}}}{\left({u}\right)}}\right)}={\frac{{{1}}}{{{2}}}}{\left({\frac{{{5}}}{{{2}}}}{\ln}{\left|{\left({y}-{\frac{{{1}}}{{{2}}}}\right)}^{{2}}+{1}\right|}+{8}{{\tan}^{{-{1}}}{\left({y}-{\frac{{{1}}}{{{2}}}}\right)}}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{4}}}}{\left({5}{\ln}{\left|{y}^{{2}}-{y}+{\frac{{{5}}}{{{4}}}}\right|}+{16}{{\tan}^{{-{1}}}{\left({y}-{\frac{{{1}}}{{{2}}}}\right)}}\right)}+{C}\)
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