The value of {d x}{7+6 x-x²} is equal to

Correct Answer is : 18log⁡(1+x7−x)+c11/8 ≤ft(1+x/7-x)+c₁81log(7−x1+x)+c1 (where c is a constant of integration)

1/4 ≤ft(1+x/7-x)+c, (where c is a constant of integration)

1/8 ≤ft(7-x/1+x)+c_{,}( where c is a constant of integration)

1/4 ≤ft(7-x/1+x)+c, (where c is a constant of integration)

1/8 ≤ft(1+x/7-x)+c₁ (where c is a constant of integration)

Correct Answer is : 18log⁡(1+x7−x)+c11/8 ≤ft(1+x/7-x)+c₁81log(7−x1+x)+c1 (where c is a constant of integration)

Test Index

MHTCET 15 May 2024 Shift 1 Paper

Question : 103 of 150

Marks: +1, -0

The value of

\int \frac{d x}{7+6 x-x^{2}}

is equal to

$\frac{1}{4} \log \left(\frac{1+x}{7-x}\right)+c$ , (where c is a constant of integration)
$\frac{1}{8} \log \left(\frac{7-x}{1+x}\right)+c_{,}$ ( where c is a constant of integration)
$\frac{1}{4} \log \left(\frac{7-x}{1+x}\right)+c$ , (where c is a constant of integration)
$\frac{1}{8} \log \left(\frac{1+x}{7-x}\right)+c_{1}$ (where c is a constant of integration)

Solution:

Concept:

Use the standard integral formula

\int \frac{dx}{a^2 - u^2} = \frac{1}{2a}\log\left|\frac{a+u}{a-u}\right| + c

Explanation:

Rewrite the denominator by completing the square.

7+6x-x^2 = 16 - (x^2 - 6x + 9) = 16 - (x-3)^2

Substitute

a = 4

and

u = x-3

Then the integral becomes

\int \frac{dx}{4^2 - (x-3)^2}

Apply the formula:

\frac{1}{2(4)} \log\left|\frac{4+(x-3)}{4-(x-3)}\right| + c

Simplify:

\frac{1}{8} \log\left|\frac{1+x}{7-x}\right| + c

Answer:

\frac{1}{8} \log\left(\frac{1+x}{7-x}\right)+c

, which matches option D.

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