Question:
Find: \[
\int \frac{x^2}{(x^2-1)(x^2+4)}\,dx
\]
Find: \[
\int \frac{x^2}{(x^2-1)(x^2+4)}\,dx
\]
Show Hint
Using a temporary algebraic substitution like \(x^2 = t\) helps you avoid handling complex terms while finding partial fractions. Just remember to convert back to \(x^2\) before applying your integration formulas!
Updated On: Jul 8, 2026
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Solution and Explanation
Concept:
This integration can be solved using the method of partial fractions. Since only \( x^2 \) appears in the integrand, we can temporarily substitute \( x^2 = t \) to find the partial fraction decomposition cleanly before substituting back and integrating.
• Standard Integral 1: \( \int \frac{1}{x^2 - a^2} dx = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C \).
• Standard Integral 2: \( \int \frac{1}{x^2 + a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C \).
Step 1: Using a temporary substitution to perform partial fraction decomposition.
Let the integrand be: \[ \frac{x^2}{(x^2 - 1)(x^2 + 4)} \] Let \( x^2 = t \) temporarily. The expression becomes: \[ \frac{t}{(t - 1)(t + 4)} = \frac{A}{t - 1} + \frac{B}{t + 4} \] Combine the fractions on the right side: \[ t = A(t + 4) + B(t - 1) \]
Step 2: Solving for the partial fraction constants \(A\) and \(B\).
To find \(A\), set \( t = 1 \): \[ 1 = A(1 + 4) + B(0) \quad \Rightarrow \quad 5A = 1 \quad \Rightarrow \quad A = \frac{1}{5} \] To find \(B\), set \( t = -4 \): \[ -4 = A(0) + B(-4 - 1) \quad \Rightarrow \quad -5B = -4 \quad \Rightarrow \quad B = \frac{4}{5} \] Substitute \(A\) and \(B\) back into the expression: \[ \frac{t}{(t - 1)(t + 4)} = \frac{1}{5(t - 1)} + \frac{4}{5(t + 4)} \]
Step 3: Replacing \(t\) back with \(x^2\) and integrating.
Now substitute \( t = x^2 \) back into our fractions: \[ \int \frac{x^2}{(x^2 - 1)(x^2 + 4)} dx = \frac{1}{5}\int \frac{1}{x^2 - 1} dx + \frac{4}{5}\int \frac{1}{x^2 + 4} dx \] Apply the standard integration formulas: \[ \frac{1}{5} \left( \frac{1}{2(1)} \log\left|\frac{x-1}{x+1}\right| \right) + \frac{4}{5} \left( \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) \right) + C \] Simplify the coefficients: \[ = \frac{1}{10}\log\left|\frac{x-1}{x+1}\right| + \frac{2}{5}\tan^{-1}\left(\frac{x}{2}\right) + C \]
• Standard Integral 1: \( \int \frac{1}{x^2 - a^2} dx = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + C \).
• Standard Integral 2: \( \int \frac{1}{x^2 + a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C \).
Step 1: Using a temporary substitution to perform partial fraction decomposition.
Let the integrand be: \[ \frac{x^2}{(x^2 - 1)(x^2 + 4)} \] Let \( x^2 = t \) temporarily. The expression becomes: \[ \frac{t}{(t - 1)(t + 4)} = \frac{A}{t - 1} + \frac{B}{t + 4} \] Combine the fractions on the right side: \[ t = A(t + 4) + B(t - 1) \]
Step 2: Solving for the partial fraction constants \(A\) and \(B\).
To find \(A\), set \( t = 1 \): \[ 1 = A(1 + 4) + B(0) \quad \Rightarrow \quad 5A = 1 \quad \Rightarrow \quad A = \frac{1}{5} \] To find \(B\), set \( t = -4 \): \[ -4 = A(0) + B(-4 - 1) \quad \Rightarrow \quad -5B = -4 \quad \Rightarrow \quad B = \frac{4}{5} \] Substitute \(A\) and \(B\) back into the expression: \[ \frac{t}{(t - 1)(t + 4)} = \frac{1}{5(t - 1)} + \frac{4}{5(t + 4)} \]
Step 3: Replacing \(t\) back with \(x^2\) and integrating.
Now substitute \( t = x^2 \) back into our fractions: \[ \int \frac{x^2}{(x^2 - 1)(x^2 + 4)} dx = \frac{1}{5}\int \frac{1}{x^2 - 1} dx + \frac{4}{5}\int \frac{1}{x^2 + 4} dx \] Apply the standard integration formulas: \[ \frac{1}{5} \left( \frac{1}{2(1)} \log\left|\frac{x-1}{x+1}\right| \right) + \frac{4}{5} \left( \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) \right) + C \] Simplify the coefficients: \[ = \frac{1}{10}\log\left|\frac{x-1}{x+1}\right| + \frac{2}{5}\tan^{-1}\left(\frac{x}{2}\right) + C \]
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