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Integration of 7+logx/(8+logx)^2 .dx please get me the ans as quick as possible

Integration of 7+logx/(8+logx)^2 .dx please get me the ans as quick as possible

Grade:12

3 Answers

Nandana
110 Points
6 years ago
hi ,
     while doing I troubled to get the integration of ex / x , if you know this please inform me
 I’ll try to get the answer … !
Sanket
36 Points
6 years ago
\int (7+log x)/(8+logx)^{2}
substutute x=e^t
dx = e^tdt
this gives
\int (7+t)/(8+t)^2 e^tdt
=\int e^t [1/(8+t)-1/(8+t)^2]
and /int e^x {f(x) + f’(x)} . dx = f(x)e^x
so, we get e^t/(8+t)
then replace t = log x to get final answer
Nandana
110 Points
6 years ago
Hi... For this question integrate ∫(7+logx/(8+logx)^2 .dx ) = ∫(7+1-1+log x)/(8+log x)^2 = ∫1/(8+log x) dx -∫1/(8+log x)^2 dx By using integration by parts u = 1/(8+log x ) ; dv = dx du = -1/(8+log x)*1/x ; v = x For the 1st part ∫1/(8+log x) dx = ×/(8+logx) +∫1/(8+log x)^2 dx Finally ,= ×/(8+logx) +∫1/(8+log x)^2 dx -∫1/(8+log x)^2 dx =x /(8+ log x) + K Thank you

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