If ∫ 1 ( x - 2 ) ^ 5 ( x - 1 ) ^ 4 dx = ∑ _ r = - 4 ^ - 1 A _ r ((x - 2)/(x - 1))r + ∑ _ r = 1 ^ 3 A _ r ((x - 2)/(x - 1))r + Bf(x), ~then f(x) =

Correct answer: A. ( x - 2 ) - ( x - 1 ) I = ∫ (1)/((x - 2)5 . (x - 1)4)dx ⇒ I = ∫ (dx)/( (x - 2)9 . ((x - 1)/(x - 2))4) Let (x - 1)/(x - 2) = t ⇒ ((x - 2) . 1 - (x - 1) . 1)/( (x - 2)2)dx = dt ⇒ (- 1)/((x - 2)2) dx = dt . . . (1) Also, (x - 1)/(x - 2) = t ⇒ x - 1 = tx - 2t ⇒ x = (2t - 1)/(t - 1) ⇒ x - 2 = (1)/(t - 1) . . . (2) ∴ I = ∫ (- dt)/((1)/((t - 1)7)) t^ 4 ⇒ I = ∫ ( (1 - t)7)/(t4) dt ⇒ I = ∫ 1 - c1 7 t + c2 7 t^ 2 - c3 7 t^ 3 + . . . . . . + c6 7 . t^ 6 - c7 7 t^ 7 t4 dt ⇒ I = ∫ (1)/(t4) dt - c1 7 ∫ (dt)/(t3) + c2 7 ∫ (dt)/(t2) - c3 7 ∫ (1)/(t) dt + . . . . - c7 7 ∫ t^ 3 dt ⇒ I = (t- 4 + 1)/(- 4 + 1) -7(t- 3 + 1)/(- 3 + 1) + c2 7 (t- 2 + 1)/(- 2 + 1) - c3 7 |t| + . . . - (t4)/(4) + c ~ ⇒ I = (- 1)/(3)((x - 1)/(x - 2))- 3 + (7)/(2)((x - 1)/(x - 2))- 2 - c3 7 | (x - 1)/(x - 2)| + . . . . . - (1)/(4)((x - 1)/(x - 2)) 4 + c ⇒ I = (- 1)/(3) ((x - 2)/(x - 1)) 3 + (7)/(2) ((x - 2)/(x - 1)) 2 + c2 7 ((x - 2)/(x - 1)) + c4 7 ((x - 2)/(x - 1))- 1 + c3 7 [ (x - 2) - ( x - 1)] . . . . - (1)/(4) ((x - 2)/(x - 1))-4 + c . . . (1) ∵ Given that ∫ 1 ( x - 2 ) ^ 5 ( x - 1 ) ^ 4 dx = ∑ _ r = - 4 ^ - 1 A _ r ((x - 2)/(x - 1))r + ∑ _ r = 1 ^ 3 A _ r ((x - 2)/(x - 1)) r + Bf (x) ∴ We get f ( x) = ( x - 2) - ( x - 1)

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main · Mathematics · Indefinite Integration

If ∫ 1 ( x - 2 ) ^ 5 ( x - 1 ) ^ 4 dx = ∑ _ r = - 4 ^ - 1 A _ r ((x...

∫ \dfrac{1}{( x - 2{)}^{5} ( x - 1{)}^{4}}dx = \mathop{∑}_{r = - 4}^{- 1}{A}_{r}\left(\dfrac{x - 2}{x - 1}\right)r + \mathop{∑}_{r = 1}^{3}{A}_{r}\left(\dfrac{x - 2}{x - 1}\right)r + Bf\left(x\right),

~then

f\left(x\right) =

Options

A $\log ( x - 2 ) - \log ( x - 1 )$
B\left(\dfrac{x - 2}{x - 1}\right) + \log x
C $x + \log \left(\dfrac{x - 2}{x - 1}\right)$
D $\log x$

Correct answer

A. $\log ( x - 2 ) - \log ( x - 1 )$

Step-by-step solution

I = ∫ \dfrac{1}{\left(x - 2\right)5 . \left(x - 1\right)4}dx $$ ⇒ I = ∫ \dfrac{dx}{ $\left(x - 2\right)9 . \left(\dfrac{x - 1}{x - 2}\right)4}$

Let $\dfrac{x - 1}{x - 2} = t$

$⇒ \dfrac{\left(x - 2\right) . 1 - \left(x - 1\right) . 1}{ \left(x - 2\right)2}dx = dt$

$⇒$ \dfrac{- 1}{\left(x - 2\right)2} $dx = dt . . .$ \left(1\right) $Also, \dfrac{x - 1}{x - 2} =$ t $$ ⇒ x - 1 = tx - 2t $$ ⇒ x = $\dfrac{2t - 1}{t - 1}$ ⇒ x - 2 = $\dfrac{1}{t - 1} . . . \left(2\right)$

$∴ I = ∫$ \dfrac{- dt}{\dfrac{1}{\left(t - 1\right)7}} $t^{4}$

$⇒ I = ∫$ \dfrac{ $\left(1 - t\right)7}{t4}$ dt $$ ⇒ I = ∫ $\dfrac{1 - c1 7 t + c2 7 t^{2} - c3 7 t^{3} + . . . . . . + c6 7 . t^{6} - c7 7 t^{7}}{t4}$ dt $$ ⇒ I = ∫ $\dfrac{1}{t4}$ dt - c1 7 ∫ $\dfrac{dt}{t3} +$ c2 7 ∫ $\dfrac{dt}{t2} -$ c3 7 ∫ $\dfrac{1}{t}$ dt + . . . . - c7 7 ∫ t^{3}dt $$ ⇒ I = $\dfrac{t- 4 + 1}{- 4 + 1} -7\dfrac{t- 3 + 1}{- 3 + 1} +$ c2 7 $\dfrac{t- 2 + 1}{- 2 + 1} -$ c3 7 \log \left|t\right| + . . . - $\dfrac{t4}{4} +$ c $~$ ⇒ I = $\dfrac{- 1}{3}\left(\dfrac{x - 1}{x - 2}\right)-$ 3 + $\dfrac{7}{2}\left(\dfrac{x - 1}{x - 2}\right)-$ 2 - c3 7 \log \left| $\dfrac{x - 1}{x - 2}\right| + . . . . . - \dfrac{1}{4}\left(\dfrac{x - 1}{x - 2}\right)$ 4 + c $$ ⇒ I = $\dfrac{- 1}{3} \left(\dfrac{x - 2}{x - 1}\right)$ 3 + $\dfrac{7}{2} \left(\dfrac{x - 2}{x - 1}\right)$ 2 + c2 7 \left(\dfrac{x - 2}{x - 1}\right) + $c4 7 \left(\dfrac{x - 2}{x - 1}\right)-$ 1 + c3 7 \left[\log \left(x - 2\right) - \log \left( $x - 1\right)\right] . . . . -$ \dfrac{1}{4} \left(\dfrac{x - 2}{x - 1}\right)-4 + c . . . \left(1\right) $$ ∵ $Given that$ ∫ \dfrac{1}{( x - 2{)}^{5} ( x - 1{)}^{4}}dx = \mathop{∑}_{r = - 4}^{- 1}{A}_{r} $\left(\dfrac{x - 2}{x - 1}\right)r + \mathop{∑}_{r = 1}^{3}{A}_{r}$ \left(\dfrac{x - 2}{x - 1}\right) $r + Bf \left(x\right)$

$∴$ We get $f \left($ x\right) = \log \left( $x - 2\right) -$ \log \left( $x - 1\right)$

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