{2 x}{4-3 x-x²} d x=·s

Correct Answer is : −85log⁡∣4+x∣−25log⁡∣1−x∣+c-8/5 |4+x|-2/5 |1-x|+c−58log∣4+x∣−52log∣1−x∣+c

Correct Answer is : −85log⁡∣4+x∣−25log⁡∣1−x∣+c-8/5 |4+x|-2/5 |1-x|+c−58log∣4+x∣−52log∣1−x∣+c

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MHTCET 11 Apr 2026 Shift 2 Paper

Question : 130 of 150

Marks: +1, -0

\int \frac{2 x}{4-3 x-x^{2}} d x=\cdots

Solution:

Concept:

Factor the denominator and use partial fractions to integrate the rational function.

Explanation:

Factor the denominator:

4-3x-x^{2} = (x+4)(1-x)

Write the integrand using partial fractions:

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

Multiply through by

(x+4)(1-x)

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

Expanding:

2x = (A+4B) + (-A+B)x

Compare coefficients:

A + 4B = 0, \quad -A + B = 2

Solving these equations gives:

A = -\frac{8}{5}, \quad B = \frac{2}{5}

Thus:

\frac{2x}{4-3x-x^{2}} = -\frac{8}{5}\cdot\frac{1}{x+4} + \frac{2}{5}\cdot\frac{1}{1-x}

Integrate term by term:

\int \frac{2x}{4-3x-x^{2}}\,dx = -\frac{8}{5}\int\frac{dx}{x+4} + \frac{2}{5}\int\frac{dx}{1-x}

Using

\int \frac{dx}{x+4} = \log|x+4|

and

\int \frac{dx}{1-x} = -\log|1-x|

= -\frac{8}{5}\log|x+4| - \frac{2}{5}\log|1-x| + C

Answer:

Option D:

-\frac{8}{5}\log|4+x| - \frac{2}{5}\log|1-x| + c

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