Let f(x)=∫ (2 x/(x2+1)(x2+3)) d x. If f(3)=(1/2)( log e 5- log e 6), then f(4) is equal to

Question

Let f(x)=∫ (2 x/(x2+1)(x2+3)) d x. If f(3)=(1/2)( log e 5- log e 6), then f(4) is equal to (A)  log e 17- log e 18 (B)  log e 19- log e 20 (C) (1/2)( log e 19- log e 17) (D) (1/2)( log e 17- log e 19)

Tardigrade Admin · Accepted Answer

Put x2=t ∫ ( dt /( t +1)( t +3))=(1/2) ∫((1/t+1)-(1/t+3)) d t f(x)=(1/2) ln ((x2+1/x2+3))+C f(3)=(1/2)( ln 10- ln 12)+C ⇒ C=0 f(4)=(1/2) ln ((17/19))

Q. Let $f (x) = \int \frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 3 )} d x$ . If $f (3) = \frac{1}{2} (lo g_{e} 5 - lo g_{e} 6)$ , then $f (4)$ is equal to

$lo g_{e} 17 - lo g_{e} 18$

$lo g_{e} 19 - lo g_{e} 20$

$\frac{1}{2} (lo g_{e} 19 - lo g_{e} 17)$

$\frac{1}{2} (lo g_{e} 17 - lo g_{e} 19)$

Put $x^{2} = t$
$\int \frac{d t}{( t + 1 ) ( t + 3 )} = \frac{1}{2} \int (\frac{1}{t + 1} - \frac{1}{t + 3}) d t$
$f (x) = \frac{1}{2} ln (\frac{x ^{2} + 1}{x ^{2} + 3}) + C$
$f (3) = \frac{1}{2} (ln 10 - ln 12) + C$
$\Rightarrow C = 0$
$f (4) = \frac{1}{2} ln (\frac{17}{19})$

Q. Let f(x)=∫(x2+1)(x2+3)2x​dx. If f(3)=21​(loge​5−loge​6), then f(4) is equal to

loge​17−loge​18

loge​19−loge​20

21​(loge​19−loge​17)

21​(loge​17−loge​19)