If ∫x+1×2+x+1dxx2+x+1=kx−1×2+x+1+C then the value of k

# If $\int \frac{x+1}{\left({x}^{2}+x+1\right)}\frac{dx}{\sqrt{{x}^{2}+x+1}}=k\frac{x-1}{\sqrt{{x}^{2}+x+1}}+C$ then the value of $k$

1. A

1/2

2. B

1/3

3. C

2/3

4. D

none of these

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### Solution:

Differentiating both the sides w.r.t.x, and simplifying , we get

$\frac{x+1}{{\left({x}^{2}+x+1\right)}^{3/2}}=\frac{3k}{2}\frac{x+1}{{\left({x}^{2}+x+1\right)}^{3/2}}⇒k=\frac{2}{3}$

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