 # 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. Armorikam 2020-11-06 Answered
Show the similarity $\int \frac{x}{{\left({x}^{2}+{a}^{2}\right)}^{m}}dx=\frac{1}{2\left(-m+1\right){\left({x}^{2}+{a}^{2}\right)}^{m-1}}+C$ using the substitution $u={x}^{2}+{a}^{2}$. Also check that the integration is correct by deriving the answer.
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Step 1
put ${x}^{2}+{a}^{2}=u$
$⇒2x\cdot dx=du$
$⇒x\cdot dx=\frac{du}{2}$
$\therefore \int \frac{x}{{\left({x}^{2}+{a}^{2}\right)}^{m}}dx=\int \left(\frac{\frac{du}{2}}{{u}^{m}}\right)\right)$
$=\frac{1}{2}\int {u}^{-m}du$
$=\frac{1}{2}\cdot \frac{{u}^{-m+1}}{-m+1}+C\left[\because \int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}\right]$
$=\frac{1}{2}\cdot \frac{{\left({x}^{2}+{a}^{2}\right)}^{-m+1}}{-m+1}+C$
$=\frac{1}{2\left(-m+1\right){\left({x}^{2}+{a}^{2}\right)}^{m-1}}+C$
Veification:
Differentiating obtained integral value with respect to x, we get
$\frac{d}{dx}\left(\frac{1}{2\left(-m+1\right){\left({x}^{2}+{a}^{2}\right)}^{m-1}}+C\right)=\frac{d}{dx}\left(\frac{{\left({x}^{2}+{a}^{2}\right)}^{-m+1}}{2\left(-m+1\right)}+C\right)$
$=\frac{1}{2\left(-m+1\right)}\cdot \left(-m+1\right){\left({x}^{2}+{a}^{2}\right)}^{-m+1-1}\cdot \frac{d\left({x}^{2}+{a}^{2}\right)}{dx}+0$
$\frac{1}{2}\cdot {\left({x}^{2}+{a}^{2}\right)}^{-m}\cdot \left(2x+0\right)$
$=\frac{x}{{\left({x}^{2}+{a}^{2}\right)}^{m}}$