Let $f(x)=\int \frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)} d x$ If $f(3)=\frac{1}{2}\left(\log _e 5-\log _e 6\right)$, then $f(4)$ is equal to

Question

Evaluate the integral: $ \int \sin^5 x \, dx $

Answer 1

To solve the integral problem and find the value of $f(4)$, we begin by evaluating the integral:

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

To simplify this, we use partial fraction decomposition:

Assume: $\frac{2x}{(x^2+1)(x^2+3)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3}$

Expression becomes:

$2x = (Ax+B)(x^2+3) + (Cx+D)(x^2+1)$

Expand and equate coefficients to solve for $A, B, C,$ and $D$:

Solving these equations:

From equation 1: $C = -A$

From equation 3: $3A - A = 2 \Rightarrow 2A = 2 \Rightarrow A = 1$

Therefore, $C = -1$

From equations 2 and 4: $D = -3B$ and $D = -B$

Equating gives: $-3B = -B \Rightarrow 2B = 0 \Rightarrow B = 0$, hence $D = 0$

With these coefficients, the partial fraction decomposition becomes:

$\frac{2x}{(x^2+1)(x^2+3)} = \frac{x}{x^2+1} - \frac{x}{x^2+3}$

Hence, the integral is:

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

For both integrals, use the substitution $u = x^2 + a$, $du = 2x \, dx$.

First integral:

$\int \frac{x}{x^2+1} \, dx = \frac{1}{2} \log_e|x^2 + 1| + C_1$

Second integral:

$\int \frac{x}{x^2+3} \, dx = \frac{1}{2} \log_e|x^2 + 3| + C_2$

Combining the results:

$f(x) = \frac{1}{2} \log_e(x^2 + 1) - \frac{1}{2} \log_e(x^2 + 3) + C$

This simplifies to:

$f(x) = \frac{1}{2} \left( \log_e\left(\frac{x^2 + 1}{x^2 + 3}\right) \right) + C$

Given $f(3) = \frac{1}{2}\left(\log_e 5 - \log_e 6\right)$, we calculate:

$\frac{1}{2} \log_e\left(\frac{10}{12}\right) + C = \frac{1}{2}\left(\log_e 5 - \log_e 6\right)$

Thus, constant $C = 0$. Therefore:

$f(x) = \frac{1}{2} \left( \log_e\left(\frac{x^2 + 1}{x^2 + 3}\right) \right)$

To find $f(4)$:

$f(4) = \frac{1}{2} \left( \log_e\left(\frac{17}{19}\right) \right)$

Hence, the answer is:

$\frac{1}{2}(\log_e 17-\log_e 19)$

The Correct Option is D