If f(x)=∫1x1/41+x1/4dx,f(0)=−6, then f(1) is equal to :

let x=t4 dx=4t3dt then ∫1x141+x14dx⇒∫4t3dtt(1+t)⇒∫4t1+tdt⇒4∫t2−1+11+tdt⇒4(t−1)+1t+1dt⇒4(t−1)22+ln(t+1)+c hence f(x)=2x14−12+4ln1+x14+cf(0)=−6⇒2+4ℓn+6=−6→C=−8 now f(1)=4ln2−8 =4(ln2−2)

If f(x)=∫1x1/41+x1/4dx,f(0)=−6, then f(1) is equal to :

let x=t4 dx=4t3dt then ∫1x141+x14dx⇒∫4t3dtt(1+t)⇒∫4t1+tdt⇒4∫t2−1+11+tdt⇒4(t−1)+1t+1dt⇒4(t−1)22+ln(t+1)+c hence f(x)=2x14−12+4ln1+x14+cf(0)=−6⇒2+4ℓn+6=−6→C=−8 now f(1)=4ln2−8 =4(ln2−2)

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If $f (x) = \int \frac{1}{x^{1 / 4} (1 + x^{1 / 4})} dx, f (0) = - 6, then f (1)$ is equal to :

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4(log_e2–2)

log_e2+2

2–log_e2

4(log_e2+2)

answer is B.

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Detailed Solution

$\begin{array}{l} let x = t^{4} \\ d x = 4 t^{3} d t \\ then \int \frac{1}{x^{\frac{1}{4}} (1 + x^{\frac{1}{4}})} d x \Rightarrow \int \frac{4 t^{3} dt}{t (1 + t)} \\ \Rightarrow \int \frac{4 t}{1 + t} dt \Rightarrow 4 \int \frac{(t^{2} - 1) + 1}{1 + t} dt \\ \Rightarrow 4 \{(t - 1) + \frac{1}{t + 1} dt \\ \Rightarrow 4 \{\frac{(t - 1)^{2}}{2} + \ln (t + 1)\} + c \end{array}$

$\begin{array}{l} hence f (x) = 2 {(x^{\frac{1}{4}} - 1)}^{2} + 4 \ln (1 + x^{\frac{1}{4}}) + c \\ f (0) = - 6 \Rightarrow 2 + 4 ℓ n + 6 = - 6 \to C = - 8 \\ now f (1) = 4 \ln 2 - 8 \\ = 4 (\ln 2 - 2) \end{array}$