# Need help proving!

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

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

Note by Akshat Sharda
2 years, 10 months ago

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

\begin{aligned} 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{aligned}

- 2 years, 10 months ago

Great solution! (+1)

- 2 years, 10 months ago

Thank you. :)

- 2 years, 10 months ago

Are you also preparing for JEE?

- 2 years, 10 months ago

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

- 2 years, 10 months ago

Really good solution! (+0!)

- 2 years, 10 months ago

Haha! :P

- 2 years, 10 months ago

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

- 2 years, 9 months ago

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

- 2 years, 9 months ago

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]

- 2 years, 9 months ago