int frac{x + 2}{x^2 - 7x + 12} dx is equal to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Step $1$: Factor the denominator: $x^2 - 7x + 12 = (x - 3)(x - 4)$.Step $2$: Use partial fractions: $\frac{x + 2}{(x - 3)(x - 4)} = \frac{A}{x - 3

Vedclass · Accepted Answer

$-5 \log |x - 3| + 6 \log |x - 4| + c$

Vedclass · Answer

(A) Step $1$: Factor the denominator: $x^2 - 7x + 12 = (x - 3)(x - 4)$.Step $2$: Use partial fractions: $\frac{x + 2}{(x - 3)(x - 4)} = \frac{A}{x - 3} + \frac{B}{x - 4}$.Step $3$: Solve for $A$ and $B$: $x + 2 = A(x - 4) + B(x - 3)$.For $x = 3$: $3 + 2 = A(3 - 4) \implies 5 = -A \implies A = -5$.For $x = 4$: $4 + 2 = B(4 - 3) \implies 6 = B \implies B = 6$.Step $4$: Integrate: $\int (\frac{-5}{x - 3} + \frac{6}{x - 4}) dx = -5 \log |x - 3| + 6 \log |x - 4| + c$.

$\int \frac{x + 2}{x^2 - 7x + 12} dx$ is equal to

Similar Questions

$\int \frac{(x + 1)}{x(1 + xe^x)^2} dx =$

If $\int \frac{2x+3}{(x-1)(x^2+1)} dx = \log_e {(x-1)^{\frac{5}{2}}(x^2+1)^a} - \frac{1}{2} \tan^{-1} x + A$ where $A$ is an arbitrary constant,then the value of $a$ is

Integrate the function: $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}$

$\int \frac{dx}{x(x^2+1)^3} = ?$

$\int \frac{1}{(x-2)(x^2+1)} dx=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module