Need help proving!

I=abxn1{(n2)x2+(n1)(a+b)x+nab}(x+a)2(x+b)2dxI=\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=bn1an12(a+b)I=\dfrac{b^{n-1}-a^{n-1}}{2(a+b)}.

Note by Akshat Sharda
2 years, 2 months ago

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

I=abxn1{(n2)x2+(n1)(a+b)x+nab}(x+a)2(x+b)2 dx=babaabxn1{(n2)x2+(n1)(a+b)x+nab}(x+a)2(x+b)2 dx=babaabxn1{n(x2+(a+b)x+ab)x(2x+a+b)}(x+a)2(x+b)2 dx=1baabxn1{(ba)n(x+a)(x+b)x(ba)(2x+a+b)}(x+a)2(x+b)2 dx=1baabxn1{(ba)n(x+a)(x+b)x((x+b)2(x+a)2)}(x+a)2(x+b)2 dx=1baab[n(ba)xn1(x+a)(x+b)xn{1(x+a)21(x+b)2}] dx=1baab[nxn1{1x+a1x+b}xn{1(x+a)21(x+b)2}] dx=1baabd(xn{1x+a1x+b})=1ba(xn{1x+a1x+b})ab=xn(x+a)(x+b)ab=bn1an12(a+b)\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}

Tapas Mazumdar - 2 years, 2 months ago

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Great solution! (+1)

Akshat Sharda - 2 years, 2 months ago

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Thank you. :)

Tapas Mazumdar - 2 years, 2 months ago

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@Tapas Mazumdar Are you also preparing for JEE?

Akshat Sharda - 2 years, 2 months ago

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

Tapas Mazumdar - 2 years, 2 months ago

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Really good solution! (+0!)

Harsh Shrivastava - 2 years, 2 months ago

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Haha! :P

Tapas Mazumdar - 2 years, 2 months ago

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

Jenson Oesterreicher - 2 years, 2 months 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 - 2 years, 2 months 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 years, 2 months ago

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