Dear student, ∫1axbx dxSubstitute, u = axbx, du = axln(a)bx + bxln(b)ax dx=∫1u*1uln(a) + uln(b) du =∫1u²(ln(a) + ln(b)) du=1ln(a) + ln(b)∫1u²du=1ln(a) + ln(b)*u-2+1-2+1Substituting back, u = axbx=1ln(a) + ln(b)*(axbx)-2+1-2+1=-a-xb-xln(a) + ln(b) + c Regards

Integrate 1/a^xb^x

Dear student,

\int \frac{1}{a^{x} b^{x}} d x S u b s t i t u t e, u = a^{x} b^{x}, d u = a^{x} \ln (a) b^{x} + b^{x} \ln (b) a^{x} d x = \int \frac{1}{u} * \frac{1}{u \ln (a) + u \ln (b)} d u = \int \frac{1}{u ² (\ln (a) + \ln (b))} d u = \frac{1}{\ln (a) + \ln (b)} \int \frac{1}{u ²} d u = \frac{1}{l n (a) + l n (b)} * \frac{u^{- 2 + 1}}{- 2 + 1} S u b s t i t u t i n g b a c k, u = a^{x} b^{x} = \frac{1}{l n (a) + l n (b)} * \frac{(a^{x} b^{x})^{- 2 + 1}}{- 2 + 1} = - \frac{a^{- x} b^{- x}}{l n (a) + l n (b)} + c

Regards

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Integrate 1/axbx

Integrate 1/a^xb^x