Question Statement

Evaluate the integral:

$\int \frac{x^2}{x^3+1}dx$

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Background and Explanation

This problem is solved using integration by substitution (u-substitution). The key insight is recognizing that the numerator $x^2$ is proportional to the derivative of the denominator $x^3+1$, making substitution the most efficient method.

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Solution

Let

$I=\int \frac{x^2}{x^3+1}dx$

To simplify this integral, we use the substitution method. Notice that the derivative of the denominator $x^3+1$ is $3x^2$, which differs from the numerator $x^2$ only by a constant factor.

Step 1: Choose the substitution $t=x^3+1$

Step 2: Differentiate both sides to find the relationship between $dx$ and $dt$ $(3x^2+0)dx=dt$ $3x^{2}dx=dt$

Step 3: Solve for $x^{2}dx$ $x^{2}dx=\frac{1}{3}dt$

Step 4: Substitute into the original integral. Replacing $x^3+1$ with $t$ and $x^{2}dx$ with $\frac{1}{3}dt$:

$I=\int \frac{1}{t}\cdot \frac{1}{3}dt$

Step 5: Factor out the constant and integrate $\begin{array}{rl}I& =\frac{1}{3}\int \frac{1}{t}dt\\ & =\frac{1}{3}\ln (t)+c\end{array}$

Step 6: Substitute back $t=x^3+1$ to express the answer in terms of $x$ $I=\frac{1}{3}\ln (x^3+1)+c$

Therefore, the solution is:

$\int \frac{x^2}{x^3+1}dx=\frac{1}{3}\ln (x^3+1)+c$

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Key Formulas or Methods Used

• Integration by Substitution: If $u=g(x)$, then $\int f(g(x))g^{\prime }(x)dx=\int f(u)du$
• Power Rule for Differentiation: $\frac{d}{dx}(x^n)=n{x}^{n-1}$, specifically $\frac{d}{dx}(x^3)=3x^2$
• Basic Integration: $\int \frac{1}{x}dx=\ln |x|+C$ (here applied as $\int \frac{1}{t}dt=\ln (t)+c$)
• Differential Relationship: $dt=\frac{dt}{dx}\cdot dx$

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Summary of Steps

1. Identify the substitution: Let $t=x^3+1$ (the denominator)
2. Differentiate: Compute $dt=3x^{2}dx$, which gives $x^{2}dx=\frac{1}{3}dt$
3. Rewrite the integral: Substitute to get $\frac{1}{3}\int \frac{1}{t}dt$
4. Integrate: Evaluate to $\frac{1}{3}\ln (t)+c$
5. Back-substitute: Replace $t$ with $x^3+1$ to obtain the final answer $\frac{1}{3}\ln (x^3+1)+c$