Let f(x) = int frac{16x + 24}{x^2 + 2x - 15} | vedclass

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

Question

(C) First, we decompose the integrand into partial fractions:$\frac{16x+24}{(x+5)(x-3)} = \frac{A}{x+5} + \frac{B}{x-3}$$16x+24 = A(x-3) + B(x+5)$Sett

Vedclass · Accepted Answer

$39$

Vedclass · Answer

(C) First, we decompose the integrand into partial fractions:$\frac{16x+24}{(x+5)(x-3)} = \frac{A}{x+5} + \frac{B}{x-3}$$16x+24 = A(x-3) + B(x+5)$Setting $x=3$, we get $72 = 8B \implies B=9$.Setting $x=-5$, we get $-56 = -8A \implies A=7$.Thus, $f(x) = \int (\frac{7}{x+5} + \frac{9}{x-3}) dx = 7 \log|x+5| + 9 \log|x-3| + C$.Given $f(4) = 14 \log 3$, we have $7 \log 9 + 9 \log 1 + C = 14 \log 3 + C = 14 \log 3$, which implies $C=0$.Now, calculate $f(7) = 7 \log|7+5| + 9 \log|7-3| = 7 \log 12 + 9 \log 4$.$f(7) = 7 \log(2^2 \cdot 3) + 9 \log(2^2) = 7(2 \log 2 + \log 3) + 18 \log 2 = 14 \log 2 + 7 \log 3 + 18 \log 2 = 32 \log 2 + 7 \log 3 = \log(2^{32} \cdot 3^7)$.Comparing with $\log(2^\alpha \cdot 3^\beta)$, we get $\alpha = 32$ and $\beta = 7$.Therefore, $\alpha + \beta = 32 + 7 = 39$.

Let $f(x) = \int \frac{16x + 24}{x^2 + 2x - 15} dx$. If $f(4) = 14 \log_e(3)$ and $f(7) = \log_e(2^\alpha \cdot 3^\beta)$, where $\alpha, \beta \in N$, then $\alpha + \beta$ is equal to:

Similar Questions

Integrate the rational function: $\frac{5x}{(x+1)(x^2-4)}$

$\int \frac{d x}{\sin x+\sin 2 x}=$

Find $\int \frac{x^{2}+1}{x^{2}-5 x+6} d x$

If $\int {\frac{{(2{x^2} + 1)\,dx}}{{({x^2} - 4)({x^2} - 1)}} = \log \left[ {{{\left( {\frac{{x + 1}}{{x - 1}}} \right)}^a}\,{{\left( {\frac{{x - 2}}{{x + 2}}} \right)}^b}} \right]} + C,$ then the values of $a$ and $b$ are respectively

If $\int \frac{x+3}{(x-1)^2(2 x-1)} d x=\frac{A}{x-1}+B \log (2 x-1)+C \log (x-1)+K$, then $A+B+C=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module