\int\frac{6x^2-17x-5}{(x+3)(x-2)^2}\dx=

Correct Answer is : log(x−2)8(x−3)2−15x−2+c

{\log}\frac{(x-2)^2}{(x-3)^4}+3/{x-2}+c

\log\{(x-2)^4(x-3)^2\}3/{x-2}+c

{\log}\frac{(x-2)^8}{(x-3)^2}-15/{x-2}+c

AP EAMCET Engineering 24 Apr 2017 Shift 1 Paper

Section: Mathematics

Question : 75 of 160

Marks: +1, -0

∫

6x2−17x−5

(x+3)(x−2)2

‌dx=

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