How do you use substitution to integrate #(5x+10)/(3x^(2) +12x - 7)#?

Answer 1

Start by factoring the 5 out of the numerator.

#int (5x+10)/(3x^(2) +12x - 7) dx = 5int (x+2)/(3x^2+12x-7) dx#
Now the derivative of the denominator is #6# times the numerator, so do a substitution to get a natural logarithm.
Let #u = 3x^2+12x-7# which makes #du = (6x+12) dx#
So our integral becomes: #5/6 int 1/u du = 5/6 ln abs u +C#.

And reversing the substitution gets us

#int (5x+10)/(3x^(2) +12x - 7) dx = 5/6 ln abs (3x^2+12x-7) +C#
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Answer 2

To integrate ( \frac{5x + 10}{3x^2 + 12x - 7} ) using substitution, let ( u = 3x^2 + 12x - 7 ). Then, find ( du ) by differentiating ( u ) with respect to ( x ). After finding ( du ), express ( \frac{5x + 10}{3x^2 + 12x - 7} ) in terms of ( u ) and ( du ). This will transform the integral into a simpler form that can be integrated more easily. Finally, integrate the expression in terms of ( u ), and substitute back the original variable ( x ) to obtain the final result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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