Question

Integrate the following with respect to x : 2x-3÷x^2+4x-12​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$\int {\frac{2x-3}{x^{2} +4x-12} } \, dx$ =  $\frac{1}{4}$ * log (x+6)⁹ (x-2)

Step-by-step explanation:

Given that $\int {\frac{2x-3}{x^{2} +4x-12} } \, dx$

We factorize the denominator that is  x² + 4x - 12

We know that 6 + (-2) = 4 and 6*(-2) = -12

Therefore, by mid term splitting, we get,

x² + 4x - 12 = x² + 6x - 2x - 12

                  = x(x+6) -2(x+6)

                  = (x+6) (x - 2)

=>   $\int {\frac{2x-3}{x^{2} +4x-12} } \, dx$ =  $\int {\frac{2x-3}{(x+6)(x-2)} } \, dx$

Now as the denominator as two not repeated factors,

We integrate using partial fractions,

=> $\frac{2x-3}{(x+6)(x-2)}$ = $\frac{A}{x+6}$ + $\frac{B}{x-2}$

                    = $\frac{A(x-2) + B(x+6)}{(x+6)(x-2)}$

=> 2x - 3 = Ax - 2A + Bx + 6B

Equating the terms on both the sides, we get,

2x = Ax - Bx  and -3 = -2A + 6B

=> 2 = A - B      and  - 3 = -2A + 6B

=> A = 2 + B

We will substitute this value in the other equation,

=>  - 3 = -2(2+B) + 6B

=> -3 = -4 -2B + 6B

=> -3+4 = 4B

=> 1 = 4B

=> B = 1/4

As A = 2+B

=> A = 2 + 1/4 = 9/4

We substitute these values in the equation.

=> $\frac{2x-3}{(x+6)(x-2)}$ = $\frac{A}{x+6}$ + $\frac{B}{x-2}$ = $\frac{9}{4(x+6)}$ + $\frac{1}{4(x-2)}$

Integrating now,

$\int {\frac{2x-3}{(x+6)(x-2)} } \, dx$ = $\int {\frac{9}{4(x+6)} } \, dx$ + $\int {\frac{1}{4(x-2)} } \, dx$

                       = $\frac{9}{4}\int {\frac{1}{(x+6)} } \, dx$ +   $\frac{1}{4}\int {\frac{1}{(x-2)} } \, dx$

                      = $\frac{9}{4}$ log |x+6| +    $\frac{1}{4}$ log |x-2|

                      = $\frac{1}{4}$ * (9log |x+6| + log |x-2|)

                      = $\frac{1}{4}$ * (log |x+6|⁹ + log |x-2|)      (as log mⁿ = n log m)

                      = $\frac{1}{4}$ * log (x+6)⁹ (x-2)                (as log m + log n = log mn)

Therefore, $\int {\frac{2x-3}{x^{2} +4x-12} } \, dx$ =  $\frac{1}{4}$ * log (x+6)⁹ (x-2)