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If dx(x+1)2x2+3x+4

=Klogf(x)+C, then 

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a
K=−13,f(x)=5−x23(x+1)+5−x23(x+1)2+2326
b
K=13f(x)=4−x(x+1)+4−x(x+1)2+2316
c
K=2,f(x)=y+y2+1,y=5−xx+1
d
none of these

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detailed solution

Correct option is D

Putting x+1=1/y, the given integral can be written as −∫dy3y2−y+2=−13∫dy(y−1/6)2+(23/6)2=−13log⁡(y−1/6)+(y−1/6)2+(23/6)2+C=−13log⁡5−x6(x+1)+5−x6(x+1)2+2362+C.


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