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\[I=\displaystyle \int^{b}_{a}\dfrac{x^{n-1}\{(n-2)x^2+(n-1)(a+b)x+nab\}}{(x+a)^2(x+b)^2}dx\]

Prove that \(I=\dfrac{b^{n-1}-a^{n-1}}{2(a+b)}\).

Note by Akshat Sharda
1 month ago

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Observe that \((x+b)^2 - (x+a)^2 = (b-a)(2x+a+b)\). Now

\[\begin{align} I &= \int_a^b \dfrac{x^{n-1} \left\{ (n-2)x^2 + (n-1)(a+b)x + nab \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{b-a}{b-a} \cdot \int_a^b \dfrac{x^{n-1} \left\{ (n-2)x^2 + (n-1)(a+b)x + nab \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{b-a}{b-a} \cdot \int_a^b \dfrac{x^{n-1} \left\{ n(x^2 + (a+b)x + ab) - x(2x+a+b) \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \dfrac{x^{n-1} \left\{ (b-a)n(x+a)(x+b) - x(b-a)(2x+a+b) \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \dfrac{x^{n-1} \left\{ (b-a)n(x+a)(x+b) - x((x+b)^2 - (x+a)^2) \right\}}{(x+a)^2 (x+b)^2} \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \left[ \dfrac{n(b-a)x^{n-1}}{(x+a)(x+b)} - x^n \left\{ \dfrac{1}{(x+a)^2} - \dfrac{1}{(x+b)^2} \right\} \right] \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \left[ nx^{n-1} \left\{ \dfrac{1}{x+a} - \dfrac{1}{x+b} \right\} - x^n \left\{ \dfrac{1}{(x+a)^2} - \dfrac{1}{(x+b)^2} \right\} \right] \ \mathrm{d}x \\ &= \dfrac{1}{b-a} \cdot \int_a^b \mathrm{d} \left( x^n \left\{ \dfrac{1}{x+a} - \dfrac{1}{x+b} \right\} \right) \\ &= \dfrac{1}{b-a} \cdot \left( x^n \left\{ \dfrac{1}{x+a} - \dfrac{1}{x+b} \right\} \right) {\huge |}_a^b \\ &= \dfrac{x^n}{(x+a)(x+b)} {\huge |}_a^b \\ &= \boxed{\dfrac{b^{n-1} - a^{n-1}}{2(a+b)}} \end{align}\]

Tapas Mazumdar - 1 month ago

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WHAT THE HECK IS THAT! I know its calculus but that is a very long equation.

Jenson Oesterreicher - 3 weeks, 4 days ago

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Well for a math lover like me, I like to deal with beautiful long expressions that turn finally to a small answer.

Tapas Mazumdar - 3 weeks, 3 days ago

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Really good solution! (+0!)

Harsh Shrivastava - 1 month ago

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Haha! :P

Tapas Mazumdar - 1 month ago

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Great solution! (+1)

Akshat Sharda - 1 month ago

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Thank you. :)

Tapas Mazumdar - 1 month ago

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@Tapas Mazumdar Are you also preparing for JEE?

Akshat Sharda - 1 month ago

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@Akshat Sharda Yes. I'm studying in FIITJEE at Raipur, CG.

Tapas Mazumdar - 1 month ago

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Hello, Some of the important books for preparing JEE Maths are ML KHANNA IA MARRON for basic concepts SL Loney for Coordinate geometry
SL Loney for Trigonometry Hall knight for Higher Algebra You must also check out for [url="https://scoop.eduncle.com/jee-advanced-exam-date-notification"] </a>JEE Advanced Date 2018 [/url]

Rahul Sharma - 2 weeks, 6 days ago

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