Evaluate \int\frac{1}{(x^2+1)^2}dx

Adela Brown 2022-01-05 Answered
Evaluate \(\displaystyle\int{\frac{{{1}}}{{{\left({x}^{{2}}+{1}\right)}^{{2}}}}}{\left.{d}{x}\right.}\)

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Expert Answer

nghodlokl
Answered 2022-01-06 Author has 2098 answers
\(\displaystyle\int{\frac{{{\left.{d}{x}\right.}}}{{{\left({x}^{{2}}+{1}\right)}^{{2}}}}}\)
Set \(\displaystyle{x}\:={\tan{{\left({u}\right)}}}\) and \(\displaystyle{\left.{d}{x}\right.}={{\sec}^{{2}}{\left({u}\right)}}{d}{u}\). Then \(\displaystyle{\left({x}^{{2}}+{1}\right)}^{{2}}={\left({{\tan}^{{2}}{\left({u}\right)}}+{1}\right)}^{{2}}={{\sec}^{{4}}{\left({u}\right)}}\) and \(\displaystyle{u}={\arctan{{\left({x}\right)}}}\)
\(\displaystyle=\int{{\cos}^{{2}}{\left({u}\right)}}{d}{u}={\frac{{{1}}}{{{2}}}}\int{\cos{{\left({2}{u}\right)}}}{d}{u}+{\frac{{{1}}}{{{2}}}}\int{1}{d}{u}\)
\(\displaystyle={\frac{{{u}}}{{{2}}}}+{\frac{{{1}}}{{{4}}}}{\sin{{\left({2}{u}\right)}}}+{C}\)
\(\displaystyle={\frac{{{x}^{{2}}{\arctan{{\left({x}\right)}}}+{x}+{\arctan{{\left({x}\right)}}}}}{{{2}{x}^{{2}}+{2}}}}+{C}\)
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Laura Worden
Answered 2022-01-07 Author has 2292 answers
There is a faster way. Substitute
\(\displaystyle{x}={\tan{{\left({z}\right)}}}\ {\left.{d}{x}\right.}={{\sec}^{{2}}{\left({z}\right)}}{\left.{d}{z}\right.}\)
thus
\(\displaystyle{\left({x}^{{2}}+{1}\right)}^{{2}}={\left({{\tan}^{{2}}{\left({z}\right)}}+{1}\right)}^{{2}}={{\sec}^{{4}}{\left({z}\right)}}\ {z}={\arctan{{\left({x}\right)}}}\)
And remembering that
\(\displaystyle{\frac{{{1}}}{{{\sec}^{{2}}}}}={{\cos}^{{2}}}\)
your integral is simply
\(\displaystyle\int{{\cos}^{{2}}{\left({z}\right)}}{\left.{d}{z}\right.}\)
Which is trivial and left to you.
The tangent/secant substitution is a great technique which most of people ignore. Study it, and you will solve lots of awesome integrals!
Final result:
\(\displaystyle{\frac{{{x}^{{2}}{\arctan{{\left({x}\right)}}}+{x}+{\arctan{{\left({x}\right)}}}}}{{{2}{x}^{{2}}+{2}}}}\)
0
star233
Answered 2022-01-11 Author has 0 answers

Let us find \(\frac{d[(ax+b)/(x^2+1)^n]}{dx}\)
\(=\frac{-ax^2-2bx+a}{(x^2+1)^{n+1}} \\=\frac{-a(x^2+1)-2bx+2a}{(x^2+1)^{n+1}} \\=-\frac{a}{(x^2+1)^n}+\frac{2a-2bx}{(x^2+1)^{n+1}}\)
Integrating both sides,
\(b=0\Rightarrow2\int\frac{dx}{(x^2+1)^{n+1}}=\int\frac{dx}{(x^2+1)^n}+\frac{x}{(x^2+1)^n} \\n=1\Rightarrow2\int\frac{dx}{(x^2+1)^2}=\int\frac{dx}{x^2+1}+\frac{x}{(x^2+1)}\)

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