How do you use substitution to integrate $\frac{5 x + 10}{3 x^{2} + 12 x - 7}$ $(5x+10)/(3x^(2) +12x - 7)$ ?

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How do you use substitution to integrate $\frac{5 x + 10}{3 x^{2} + 12 x - 7}$ $(5x+10)/(3x^(2) +12x - 7)$ ?

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Answer 1 · 2015-09-06T17:25:36

Start by factoring the 5 out of the numerator.

Explanation:

$\int \frac{5 x + 10}{3 x^{2} + 12 x - 7} d x = 5 \int \frac{x + 2}{3 x^{2} + 12 x - 7} d x$ $int (5x+10)/(3x^(2) +12x - 7) dx = 5int (x+2)/(3x^2+12x-7) dx$

Now the derivative of the denominator is $6$ $6$ times the numerator, so do a substitution to get a natural logarithm.

Let $u = 3 x^{2} + 12 x - 7$ $u = 3x^2+12x-7$ which makes $d u = (6 x + 12) d x$ $du = (6x+12) dx$

So our integral becomes: $\frac{5}{6} \int \frac{1}{u} d u = \frac{5}{6} ln | u | + C$ $5/6 int 1/u du = 5/6 ln abs u +C$ .

And reversing the substitution gets us

$\int \frac{5 x + 10}{3 x^{2} + 12 x - 7} d x = \frac{5}{6} ln ∣ ∣ 3 x^{2} + 12 x - 7 ∣ ∣ + C$ $int (5x+10)/(3x^(2) +12x - 7) dx = 5/6 ln abs (3x^2+12x-7) +C$

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How do you use substitution to integrate $\frac{5 x + 10}{3 x^{2} + 12 x - 7}$ $(5x+10)/(3x^(2) +12x - 7)$ ?

1 Answer

Explanation:

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How do you use substitution to integrate $\frac{5 x + 10}{3 x^{2} + 12 x - 7}$ $(5x+10)/(3x^(2) +12x - 7)$ ?