If {d x}{x^{4}+5 x²+4}=A ^{-1} x+B ^{-1} x/2+c where c is a constant of integration, then

Correct Answer is : A=13,B=−16A=1/3, B=-1/6A=31,B=−61

Correct Answer is : A=13,B=−16A=1/3, B=-1/6A=31,B=−61

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MHTCET 26 Apr 2025 Shift 1 Paper

Question : 146 of 150

Marks: +1, -0

\int \frac{d x}{x^{4}+5 x^{2}+4}=A \tan ^{-1} x+B \tan ^{-1} \frac{x}{2}+c

where c is a constant of integration, then

Solution:

Concept:

Use partial fractions to split the integrand, then integrate using standard inverse tangent formulas.

Explanation:

Factor the denominator as

x^4+5x^2+4 = (x^2+1)(x^2+4)

Let

\frac{1}{(x^2+1)(x^2+4)} = \frac{A}{x^2+1} + \frac{B}{x^2+4}

Multiplying by the denominator:

1 = A(x^2+4) + B(x^2+1)

Comparing coefficients:

A+B=0

and

4A+B=1

Solving gives

A = \frac{1}{3}

B = -\frac{1}{3}

Therefore,

I = \frac{1}{3}\int \frac{dx}{x^2+1} - \frac{1}{3}\int \frac{dx}{x^2+4}

Now,

\int \frac{dx}{x^2+1} = \tan^{-1}x + c_1

and

\int \frac{dx}{x^2+4} = \frac{1}{2}\tan^{-1}\frac{x}{2} + c_2

I = \frac{1}{3}\tan^{-1}x - \frac{1}{6}\tan^{-1}\frac{x}{2} + c

Comparing with

A\tan^{-1}x + B\tan^{-1}\frac{x}{2}+c

yields

A=\frac{1}{3}

B=-\frac{1}{6}

Answer:

A=\frac{1}{3}

B=-\frac{1}{6}

, which is option B.

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