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

\(\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{1}}{{{2}{\left(-{m}+{1}\right)}{\left({x}^{{2}}+{a}^{{2}}\right)}^{{{m}-{1}}}}}+{C}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{{\left({x}^{{2}}+{a}^{{2}}\right)}^{{-{m}+{1}}}}}{{{2}{\left(-{m}+{1}\right)}}}+{C}\right)}\)

\(\displaystyle=\frac{{1}}{{{2}{\left(-{m}+{1}\right)}}}\cdot{\left(-{m}+{1}\right)}{\left({x}^{{2}}+{a}^{{2}}\right)}^{{-{m}+{1}-{1}}}\cdot\frac{{{d}{\left({x}^{{2}}+{a}^{{2}}\right)}}}{{{\left.{d}{x}\right.}}}+{0}\)

\(\displaystyle\frac{{1}}{{2}}\cdot{\left({x}^{{2}}+{a}^{{2}}\right)}^{{-{m}}}\cdot{\left({2}{x}+{0}\right)}\)

\(\displaystyle=\frac{{x}}{{{\left({x}^{{2}}+{a}^{{2}}\right)}^{{m}}}}\)

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

\(\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{1}}{{{2}{\left(-{m}+{1}\right)}{\left({x}^{{2}}+{a}^{{2}}\right)}^{{{m}-{1}}}}}+{C}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{{\left({x}^{{2}}+{a}^{{2}}\right)}^{{-{m}+{1}}}}}{{{2}{\left(-{m}+{1}\right)}}}+{C}\right)}\)

\(\displaystyle=\frac{{1}}{{{2}{\left(-{m}+{1}\right)}}}\cdot{\left(-{m}+{1}\right)}{\left({x}^{{2}}+{a}^{{2}}\right)}^{{-{m}+{1}-{1}}}\cdot\frac{{{d}{\left({x}^{{2}}+{a}^{{2}}\right)}}}{{{\left.{d}{x}\right.}}}+{0}\)

\(\displaystyle\frac{{1}}{{2}}\cdot{\left({x}^{{2}}+{a}^{{2}}\right)}^{{-{m}}}\cdot{\left({2}{x}+{0}\right)}\)

\(\displaystyle=\frac{{x}}{{{\left({x}^{{2}}+{a}^{{2}}\right)}^{{m}}}}\)