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Q.Evaluate ∫(ax−bx)2axbx dx\int \frac{(a^x - b^x)^2}{a^x b^x}\, dx, (a>0,a≠1,b>0,b≠1)(a > 0, a \neq 1, b > 0, b \neq 1) on RR.

Telangana TsbieTelangana Board of Intermediate Education 2022Subjective· 7mImportance★★★★★
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Expand the square, divide each term by axbxa^xb^x, and integrate the resulting exponential terms.

Expand the numerator:

(ax−bx)2=a2x−2axbx+b2x(a^x-b^x)^2=a^{2x}-2a^xb^x+b^{2x}

Divide by axbxa^xb^x:

a2x−2axbx+b2xaxbx=axbx−2+bxax=(ab)x+(ba)x−2\frac{a^{2x}-2a^xb^x+b^{2x}}{a^xb^x}=\frac{a^x}{b^x}-2+\frac{b^x}{a^x}=\left(\frac{a}{b}\right)^x+\left(\frac{b}{a}\right)^x-2

Integrate term by term, using ∫kx dx=kxlog⁡k+c\int k^x\,dx=\dfrac{k^x}{\log k}+c: …

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