Solved: "Show the similarity ∫x(x2+a2)mdx=12(−m+1)(x2+a2)m−1+C using the substitution u=x2+a2...."

Armorikam

Answered question

2020-11-06

Show the similarity

\int \frac{x}{{(x^{2} + a^{2})}^{m}} d x = \frac{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.

escumantsu

Skilled2020-11-07Added 98 answers

Step 1
put

x^{2} + a^{2} = u

\Rightarrow 2 x \cdot d x = d u

\Rightarrow x \cdot d x = \frac{d u}{2}

∴ \int \frac{x}{{(x^{2} + a^{2})}^{m}} d x = \int (\frac{\frac{d u}{2}}{u^{m}}))

= \frac{1}{2} \int u^{- m} d u

= \frac{1}{2} \cdot \frac{u^{- m + 1}}{- m + 1} + C [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1}]

= \frac{1}{2} \cdot \frac{{(x^{2} + a^{2})}^{- m + 1}}{- m + 1} + C

= \frac{1}{2 (- m + 1) {(x^{2} + a^{2})}^{m - 1}} + C

Veification:
Differentiating obtained integral value with respect to x, we get

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

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

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

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

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