f(x) = 1x^1/4(1+x^1/4),dx, f(0)=-6.

Question

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

Accepted Answer

Let $x=t^4$: $f(x) = \displaystyle\int\frac{4t^3\,dt}{t(1+t)} = 4\int\frac{t^2}{1+t}\,dt = 4\int\left[(t-1)+\frac1{t+1}\right]dt = 2(t-1)^2+4\ln(t+1)+C$

With $t=x^{1/4}$: $f(x) = 2(x^{1/4}-1)^2+4\ln(1+x^{1/4})+C$. Using $f(0)=-6$: $2+C=-6 \implies C=-8$.

$$f(1) = 0+4\ln2-8 = 4(\ln2-2)$$

Solutions · 1