Question Statement

Evaluate the integral: $\int \frac{z^3}{1+z^4}dz$

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Background and Explanation

This problem demonstrates the substitution method for integration. Notice that the numerator $z^3$ is proportional to the derivative of the inner function $z^4$ appearing in the denominator, which suggests using $u$-substitution with $t=z^4$.

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Solution

We begin by denoting the integral as $I$: $I=\int \frac{z^3}{1+z^4}dz$

To solve this, we use the substitution method. Let: $t=z^4$

Differentiating both sides with respect to $z$: $\frac{dt}{dz}=4z^3$

Using differentials, we can write this relationship as: $\begin{array}{rl}4z^{3}dz& =dt\\ z^{3}dz& =\frac{1}{4}dt\end{array}$

Now substitute $t=z^4$ and $z^{3}dz=\frac{1}{4}dt$ into equation (1): $\begin{array}{rl}I& =\int \frac{\frac{1}{4}dt}{1+t}\\ & =\frac{1}{4}\int \frac{1}{1+t}dt\end{array}$

This reduces to the standard logarithmic form. Integrating with respect to $t$: $I=\frac{1}{4}\ln (1+t)+c$

(Note: Since $z^4\ge 0$ for all real $z$, the expression $1+z^4$ is always positive, so we can write $\ln (1+t)$ without absolute value bars.)

Finally, substitute back $t=z^4$ to express the solution in terms of the original variable $z$: $I=\frac{1}{4}\ln \left(1+z^4\right)+c$

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Key Formulas or Methods Used

• Substitution Rule (u-substitution): $\int f(g(z))g^{\prime }(z)dz=\int f(t)dt$ where $t=g(z)$
• Power Rule for Differentiation: $\frac{d}{dz}(z^n)=n{z}^{n-1}$ (specifically $\frac{d}{dz}(z^4)=4z^3$)
• Logarithmic Integration: $\int \frac{1}{x}dx=\ln |x|+C$ (applied here as $\int \frac{1}{1+t}dt=\ln (1+t)+c$ since $1+t>0$)

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Summary of Steps

1. Identify the integral and set $I=\int \frac{z^3}{1+z^4}dz$
2. Choose substitution $t=z^4$ because the numerator $z^3$ is proportional to the derivative of $z^4$
3. Compute the differential $dt=4z^{3}dz$ and solve for $z^{3}dz=\frac{1}{4}dt$
4. Rewrite the integral in terms of $t$: $\frac{1}{4}\int \frac{1}{1+t}dt$
5. Integrate to obtain $\frac{1}{4}\ln (1+t)+c$
6. Back-substitute $t=z^4$ to get the final answer: $\frac{1}{4}\ln (1+z^4)+c$