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)=3 3x+6, where a= 3, b=6 So, nbsp;I=∫3 3x+6dx= 3ln( 3x+6)3+C ⇒ I= ...

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Byju's Answer

Standard XII

Physics

Antiderivative

Evaluate ∫3-3...

Question

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

ln (- 3 x + 6)

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- ln (- 3 x + 6)

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

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

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Solution

The correct option is B $- ln (- 3 x + 6)$

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,
$f (x) = \frac{3}{- 3 x + 6}$ , where $a = - 3, b = 6$
So, $\int \frac{3}{- 3 x + 6} d x = \frac{- 3 ln (- 3 x + 6)}{3} + C$
$\Rightarrow - ln (- 3 x + 6)$

Suggest Corrections

Evaluate ∫3−3x+6dx

The correct option is B −ln(−3x+6) 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)=3−3x+6, where a=−3,b=6 So, ∫3−3x+6dx=−3ln(−3x+6)3+C ⇒ −ln(−3x+6)

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