Let f(x)=∫2x(x2+1)(x2+3) dx . If f(3)=12(loge5−loge6) , then f(4) is equal to

f(x)=∫2x(x2+1)(x2+3)dxLet x2=t2xdx=dt∫dt(t+1)(t+3)=12∫(t+3)−(t+1)(t+1)(t+3)dt=12[ln|t+1|−ln|t+3|]+C2=12[ln|x2+1|−ln|x2+3|]+C2∴f(3)=12[ln5−ln6]∴12[ln5−ln6]=12[ln10−ln12]+C2⇒C=0∴f(x)=12[ln|x2+1|−ln|x2+3|]f(4)=12[ln17−ln19]

Let f(x)=∫2x(x2+1)(x2+3) dx . If f(3)=12(loge5−loge6) , then f(4) is equal to

f(x)=∫2x(x2+1)(x2+3)dxLet x2=t2xdx=dt∫dt(t+1)(t+3)=12∫(t+3)−(t+1)(t+1)(t+3)dt=12[ln|t+1|−ln|t+3|]+C2=12[ln|x2+1|−ln|x2+3|]+C2∴f(3)=12[ln5−ln6]∴12[ln5−ln6]=12[ln10−ln12]+C2⇒C=0∴f(x)=12[ln|x2+1|−ln|x2+3|]f(4)=12[ln17−ln19]

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

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$\log_{e} 19 - \log_{e} 20$

$\log_{e} 17 - \log_{e} 18$

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

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

answer is D.

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Detailed Solution

$\begin{array}{l} f (x) = \int \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} dx \\ Let x^{2} = t \\ 2 xdx = dt \\ \int \frac{dt}{(t + 1) (t + 3)} \\ = \frac{1}{2} \int \frac{(t + 3) - (t + 1)}{(t + 1) (t + 3)} dt \\ = \frac{1}{2} [\ln | t + 1 | - \ln | t + 3 |] + \frac{C}{2} \\ = \frac{1}{2} [\ln | x^{2} + 1 | - \ln | x^{2} + 3 |] + \frac{C}{2} \\ ∴ f (3) = \frac{1}{2} [\ln 5 - \ln 6] \\ ∴ \frac{1}{2} [\ln 5 - \ln 6] = \frac{1}{2} [\ln 10 - \ln 12] + \frac{C}{2} \\ \Rightarrow C = 0 \\ ∴ f (x) = \frac{1}{2} [\ln | x^{2} + 1 | - \ln | x^{2} + 3 |] \\ f (4) = \frac{1}{2} [\ln 17 - \ln 19] \end{array}$