Show the similarity int(x)/((x^2+a^2)^m)dx=(1)/(2(-m+1)(x^2+a^2)^(m-1))+C using the substitution u = x^2 + a^2. Also check that the integration is correct by deriving the answer.

Show the similarity int(x)/((x^2+a^2)^m)dx=(1)/(2(-m+1)(x^2+a^2)^(m-1))+C using the substitution u = x^2 + a^2. Also check that the integration is correct by deriving the answer.

Question
Similarity
asked 2020-11-06
Show the similarity \(\displaystyle\int\frac{{{x}}}{{{\left({x}^{{2}}+{a}^{{2}}\right)}^{{m}}}}{\left.{d}{x}\right.}=\frac{{{1}}}{{{2}{\left(-{m}+{1}\right)}{\left({x}^{{2}}+{a}^{{2}}\right)}^{{{m}-{1}}}}}+{C}\) using the substitution \(\displaystyle{u}={x}^{{2}}+{a}^{{2}}\). Also check that the integration is correct by deriving the answer.

Answers (1)

2020-11-07
Step 1
put \(\displaystyle{x}^{{2}}+{a}^{{2}}={u}\)
\(\displaystyle\Rightarrow{2}{x}\cdot{\left.{d}{x}\right.}={d}{u}\)
\(\displaystyle\Rightarrow{x}\cdot{\left.{d}{x}\right.}=\frac{{{d}{u}}}{{2}}\)
\(\displaystyle\therefore\int\frac{{{x}}}{{{\left({x}^{{2}}+{a}^{{2}}\right)}^{{m}}}}{\left.{d}{x}\right.}=\int{\left(\frac{{\frac{{{d}{u}}}{{2}}}}{{u}^{{m}}}\right)}{)}\)
\(\displaystyle=\frac{{1}}{{2}}\int{u}^{{-{{m}}}}{d}{u}\)
\(\displaystyle=\frac{{1}}{{2}}\cdot\frac{{{u}^{{-{m}+{1}}}}}{{-{m}+{1}}}+{C}{\left[\because\int{x}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}^{{{n}+{1}}}}}{{{n}+{1}}}\right]}\)
\(\displaystyle=\frac{{1}}{{2}}\cdot\frac{{{\left({x}^{{2}}+{a}^{{2}}\right)}^{{-{m}+{1}}}}}{{-{m}+{1}}}+{C}\)
\(\displaystyle=\frac{{1}}{{{2}{\left(-{m}+{1}\right)}{\left({x}^{{2}}+{a}^{{2}}\right)}^{{{m}-{1}}}}}+{C}\)
Veification:
Differentiating obtained integral value with respect to x, we get
\(\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{1}}{{{2}{\left(-{m}+{1}\right)}{\left({x}^{{2}}+{a}^{{2}}\right)}^{{{m}-{1}}}}}+{C}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{{\left({x}^{{2}}+{a}^{{2}}\right)}^{{-{m}+{1}}}}}{{{2}{\left(-{m}+{1}\right)}}}+{C}\right)}\)
\(\displaystyle=\frac{{1}}{{{2}{\left(-{m}+{1}\right)}}}\cdot{\left(-{m}+{1}\right)}{\left({x}^{{2}}+{a}^{{2}}\right)}^{{-{m}+{1}-{1}}}\cdot\frac{{{d}{\left({x}^{{2}}+{a}^{{2}}\right)}}}{{{\left.{d}{x}\right.}}}+{0}\)
\(\displaystyle\frac{{1}}{{2}}\cdot{\left({x}^{{2}}+{a}^{{2}}\right)}^{{-{m}}}\cdot{\left({2}{x}+{0}\right)}\)
\(\displaystyle=\frac{{x}}{{{\left({x}^{{2}}+{a}^{{2}}\right)}^{{m}}}}\)
0

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