_0^4 0.05em 11+ x dx=

Question

Accepted Answer

Correct answer: C. (c) : Let I= _0^4 0.05em 11+ x dx Put 1+ x=t ⇒ 12 x dx= dt ⇒ dx=2 x dt=2 (t-1 ) dt When x=0 ,t=1 and when x=4 ,t=3 ∴ I= _1^3 0.05em 2 (t-1 )t dt= _1^32 dt- _1^3 0.05em 2t dt=2 [t ]_1^3-2 [ log ⁡t ]_1^3 =2 2-2 ( log ⁡3- log ⁡1 )=4-2 log ⁡3=4 log ⁡e- log ⁡3^2 = log ⁡e^4- log ⁡9= log ⁡ ( e^49 ) ∴ I= log ⁡ ( e^49 )

$\int_{0}^{4} \frac{1}{1 + x} dx =$

Solution

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