JEE MAIN - Mathematics (2018 - 15th April Morning Slot - No. 19)
If $$f\left( {{{x - 4} \over {x + 2}}} \right) = 2x + 1,$$ (x $$ \in $$ R $$-$${1, $$-$$ 2}), then $$\int f \left( x \right)dx$$ is equal to :
(where C is a constant of integration)
(where C is a constant of integration)
12 loge | 1 $$-$$ x | + 3x + C
$$-$$ 12 loge | 1 $$-$$ x | $$-$$ 3x + C
12 loge | 1 $$-$$ x | $$-$$ 3x + C
$$-$$ 12 loge | 1 $$-$$ x | + 3x + C
Explanation
Let, $${{{x - 4} \over {x + 2}}}$$ = t
$$ \Rightarrow $$ x - 4 = t(x+2)
$$ \Rightarrow $$ x (1 -t) = 2(t+2)
$$ \Rightarrow $$ x = $${{2(t + 2)} \over {1 - t}}$$
$$ \therefore $$ f(t) = 2($${{2(t + 2)} \over {1 - t}}$$) + 1
= $${{4t + 8} \over {1 - t}} + 1$$
= $${{3t + 9} \over {1 - t}}$$
= $${{3(t + 3)} \over {1 - t}}$$
= $${{3(t - 1 + 4)} \over {1 - t}}$$
= - 3 + $$12 \over 1 - t$$
$$ \therefore $$ f(x) = - 3 + $$12 \over 1 - x$$
$$ \therefore $$ $$\int {f(x)dx} $$
= $$\int {\left( { - 3 + {{12} \over {1 - x}}} \right)dx} $$
= $$-12{\log _e}|1 - x|$$ - 3x + C
$$ \Rightarrow $$ x - 4 = t(x+2)
$$ \Rightarrow $$ x (1 -t) = 2(t+2)
$$ \Rightarrow $$ x = $${{2(t + 2)} \over {1 - t}}$$
$$ \therefore $$ f(t) = 2($${{2(t + 2)} \over {1 - t}}$$) + 1
= $${{4t + 8} \over {1 - t}} + 1$$
= $${{3t + 9} \over {1 - t}}$$
= $${{3(t + 3)} \over {1 - t}}$$
= $${{3(t - 1 + 4)} \over {1 - t}}$$
= - 3 + $$12 \over 1 - t$$
$$ \therefore $$ f(x) = - 3 + $$12 \over 1 - x$$
$$ \therefore $$ $$\int {f(x)dx} $$
= $$\int {\left( { - 3 + {{12} \over {1 - x}}} \right)dx} $$
= $$-12{\log _e}|1 - x|$$ - 3x + C
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