Answer
The co-efficient of $x^3$ is $0$ and the co-efficient of $x^4$ is $0$.
Work Step by Step
The power series of $(1+x)^{-1}$ can be written as:
$1-x+x^2-x^3+......=\Sigma_{k=0}^{\infty} (-1)^k x^k$
Replace $x$ by $t^4$ in the above series to obtain:
$\int_{0}^{x} \dfrac{dt}{1+t^4} \ dt=\int_0^x (1+t^4)^{-1} \ dt \\=\int_0^x [1-t^4+(t^4)^2-(t^4)^3+......]\ dt \\=[t-\dfrac{t^5}{5}+\dfrac{t^9}{9}-\dfrac{t^{13}}{13}+......]_0^x\\=x-\dfrac{x^5}{5}+\dfrac{x^9}{9}-\dfrac{x^{13}}{13}+......$
Therefore, we find that the co-efficient of $x^3$ is $0$ and the co-efficient of $x^4$ is $0$.
