Evaluate: 10 u 9 + 10 u log e 10 u 10 + 10 u ? d u

Question

Evaluate: $\int \frac{10 u^{9} + 10^{u} \log_{e} 10}{u^{10} + 10^{u}} d u$

Smart Achievers · Accepted Answer

(1)

Ans. $\log (10^{u} + u^{10}) + C$

Explanation

$The given integral is$ $\int \frac{10 u^{9} + 10^{u} \log_{e} 10}{u^{10} + 10^{u}} d u$

$Let$ $u^{10} + 10^{u} = v$

$Differentiating both sides, we get$

$\frac{d}{d u} (u^{10} + 10^{u}) = \frac{d v}{d u}$

$10 u^{9} + 10^{u} \log_{e} 10 = \frac{d v}{d u}$

$(10 u^{9} + 10^{u} \log_{e} 10) d u = d v$

$Substitute v for u^{10} + 10^{u} and d v for (10 u^{9} + 10^{u} \log_{e} 10) d u in the given integral.$

$\int \frac{10 u^{9} + 10^{u} \log_{e} 10}{u^{10} + 10^{u}} d u = \int \frac{d v}{v}$

$= \int \frac{1}{v} d v$

$= \log v + C$

$= \log (u^{10} + 10^{u}) + C$ $(∵ v = u^{10} + 10^{u})$

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