Primitive of Reciprocal of x by x squared plus a squared squared

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Theorem

$\ds \int \frac {\d x} {x \paren {x^2 + a^2}^2} = \frac 1 {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {x^2 + a^2} } + C$


Proof

Let:

\(\ds \int \frac {\d x} {x \paren {x^2 + a^2}^2}\) \(=\) \(\ds \int \paren {\frac 1 {a^4 x} - \frac x {a^4 \paren {x^2 + a^2} } - \frac x {a^2 \paren {x^2 + a^2}^2} } \rd x\) Partial Fraction Expansion
\(\ds \) \(=\) \(\ds \frac 1 {a^4} \int \frac {\d x} x - \frac 1 {a^4} \int \frac {x \rd x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2}\) Linear Combination of Integrals
\(\ds \) \(=\) \(\ds \frac 1 {a^4} \ln x - \frac 1 {a^4} \int \frac {x \rd x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C\) Primitive of Reciprocal
\(\ds \) \(=\) \(\ds \frac 1 {a^4} \ln x - \frac 1 {a^4} \paren {\frac 1 2 \map \ln {x^2 + a^2} } - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C\) Primitive of $\dfrac x {x^2 + a^2}$
\(\ds \) \(=\) \(\ds \frac 1 {a^4} \ln x - \frac 1 {a^4} \paren {\frac 1 2 \map \ln {x^2 + a^2} } - \frac 1 {a^2} \paren {-\frac 1 {2 \paren {x^2 + a^2} } } + C\) Primitive of $\dfrac x {\left({x^2 + a^2}\right)^2}$
\(\ds \) \(=\) \(\ds \frac 1 {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {x^2 + a^2} } + C\) simplification

$\blacksquare$


Sources