As we know that ∫f(ax+b)dx=F(ax+b)a+C, where a,b and C are constants. We have here, f(x)=6 3x+6, where a= 3, b=6 So, nbsp;∫6 3x+6dx= 6ln( 3x+6)3+C ⇒ 2ln( 3 ...

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Byju's Answer

Standard XI

Physics

Antiderivative

Evaluate ∫6-3...

Question

Evaluate $\int \frac{6}{- 3 x + 6} d x$

\frac{e^{- 3 x + 6}}{2} + C

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- 2 ln (- 3 x + 6) + C

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- 2 e^{- 3 x + 6} + C

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\frac{ln (- 3 x + 6)}{2} + C

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Solution

The correct option is B $- 2 ln (- 3 x + 6) + C$
As we know that $\int f (a x + b) d x = \frac{F (a x + b)}{a} + C$ , where $a, b$ and $C$ are constants.
We have here,
$f (x) = \frac{6}{- 3 x + 6}$ , where $a = - 3, b = 6$
So, $\int \frac{6}{- 3 x + 6} d x = \frac{- 6 ln (- 3 x + 6)}{3} + C$
$\Rightarrow - 2 ln (- 3 x + 6) + C$

Suggest Corrections

Evaluate ∫6−3x+6dx

The correct option is B −2ln(−3x+6)+CAs we know that ∫f(ax+b)dx=F(ax+b)a+C, where a,b and C are constants. We have here, f(x)=6−3x+6, where a=−3,b=6 So, ∫6−3x+6dx=−6ln(−3x+6)3+C ⇒ −2ln(−3x+6)+C

Evaluate $\int \frac{6}{- 3 x + 6} d x$