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Mathematics Question on Differential equations

If ∫dx(x+2)(x2+1)=alog⁡∣1+x2∣+btan⁡−1x+15log⁡∣x+2∣+c\int \frac{dx}{(x+2)(x^2 + 1)} = a \log |1+x^2| + b \tan^{-1} x + \frac{1}{5} \log |x+2| + c, then

A

a=−110a = -\frac{1}{10} ,b=25 b = \frac{2}{5}

B

a=−110,b=−25a = \frac{-1}{10} , b = - \frac{2}{5}

C

a=110,b=25a =\frac{ 1}{10} , b = \frac{2}{5}

D

a=110,b=−25a = \frac{1}{10} , b = -\frac{ 2}{5}

Answer

a=−110a = -\frac{1}{10} ,b=25 b = \frac{2}{5}

Explanation

Solution

To find the values of a and b, we can compare the given integral expression with the expression
alog⁡∣1+x2∣+btan⁡−1(x)+15log⁡∣x+2∣+ca \log |1 + x^2| + b \tan^{-1}(x) + \frac{1}{5} \log |x + 2| + c
Comparing the integrand of the given integral with the expression
alog⁡∣1+x2∣+btan⁡−1(x)+15log⁡∣x+2∣+ca \log |1 + x^2| + b \tan^{-1}(x) + \frac{1}{5} \log |x + 2| + c,
we can see that:
a=−110andb=25a = -\frac{1}{10} \quad \text{and} \quad b = \frac{2}{5}
Therefore, option (A) a=−110andb=25a = -\frac{1}{10} \quad \text{and} \quad b = \frac{2}{5} is the correct answer.