Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

${\int }_0^1\frac{x}{1+x^4}dx$

Short answer

The solution of the given integral is ${\int }_0^1\frac{x}{1+x^4}dx=\frac{\mathrm{\pi }}{8}$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

${\int }_0^1\frac{x}{1+x^4}dx$

Step 2. Using the substitution method. 

${\int }_0^1\frac{x}{1+x^4}dx={\int }_0^1\frac{x}{1+{\left(x^2\right)}^2}dx$

Let

role="math" localid="1649086854342" $$\begin{gathered} u=x^2 \\ \frac{du}{dx}=2x \\ du=2xdx \\ \frac{1}{2}du=xdx \end{gathered}$$

Step 3. We will now write the limits of integration  in terms of the new variable u.

When $x=0$, we have

$$\begin{gathered} u=x^2 \\ u=0^2 \\ u=0 \end{gathered}$$

When role="math" localid="1649086870398" $x=1$, we have

$$\begin{gathered} u=x^2 \\ u={\left(1\right)}^{2}u=1 \end{gathered}$$

Step 4. Using the information in equations, we can change variables completely:

$$\begin{gathered} {\int }_0^1\frac{x}{1+x^4}dx=\frac{1}{2}{\int }_0^1\frac{1}{1+u^2}du \\ {\int }_0^1\frac{x}{1+x^4}dx=\frac{1}{2}{\left[{\tan }^{-1}x\right]}_0^1 \\ {\int }_0^1\frac{x}{1+x^4}dx=\frac{1}{2}\left[{\tan }^{-1}1-{\tan }^{-1}0\right] \\ {\int }_0^1\frac{x}{1+x^4}dx=\frac{1}{2}\left[\frac{\mathrm{\pi }}{4}-0^o\right] \\ {\int }_0^1\frac{x}{1+x^4}dx=\frac{1}{2}·\frac{\mathrm{\pi }}{4} \\ {\int }_0^1\frac{x}{1+x^4}dx=\frac{\mathrm{\pi }}{8} \end{gathered}$$