Q.Integrate the function $e^{3\log x}(x^4+1)^{-1}$

The key idea is to simplify the integrand using logarithm properties, then apply U Substitution.

First, simplify $e^{3\log x}$. Since $e^{\log a} = a$, we have $e^{3\log x} = x^3$. The integral becomes:

Now, let $u = x^4 + 1$. Then $du = 4x^3 \, dx$, so $x^3 \, dx = \frac{du}{4}$. Substitute:

$\int \frac{1}{u} \cdot \frac{du}{4} = \frac{1}{4} \int \frac{du}{u}$ …

Let’s start with the concept. When you see an expression like $e^{3\log x}$, your first instinct should be to simplify it using the fundamental relationship between exponentials and logarithms: $e^{\log a} = a$ for $a>0$. Here, $3\log x = \log(x^3)$, so $e^{3\log x} = e^{\log(x^3)} = x^3$. This is a classic trick — the exponential and natural log are inverse functions, so they cancel each other out, leaving just the argument.

Why does this matter? Because the original integrand looks messy, but after simplification, it becomes a rational function that screams for a simple substitution. The denominator $x^4+1$ and the numerator $x^3$ are perfectly set up for a $u$-substitution where $u = x^4+1$, because its derivative $du = 4x^3\,dx$ contains the $x^3$ factor.

Let’s work through it step by step.

1. Simplify the exponential term. Recall that $e^{a\log x} = x^a$ for $x > 0$ (since $\log x$ is defined for positive $x$). Here, $a = 3$, so:

The integrand becomes:

2. Set up the substitution. We want to integrate $\int \frac{x^3}{x^4+1}\,dx$. Notice that the derivative of the denominator $x^4+1$ is $4x^3$, which is a constant multiple of the numerator. This is the hallmark of a $u$-substitution. Let $u = x^4 + 1$. Then:

3. Rewrite the integral in terms of $u$. Substitute $x^3\,dx = du/4$ and $x^4+1 = u$:

4. Integrate with respect to $u$. The integral $\int \frac{du}{u}$ is a standard result: $\log|u| + C$. So: …

Method: Simplify $e^{a\log x}=x^{a}$, then use $\int\frac{g'}{g}=\log|g|$

Use this when an exp-log factor multiplies a rational function: convert to a power of $x$, then check whether the numerator is (a multiple of) the denominator's derivative.

Steps

Step 1: Rewrite $e^{a\log x}=x^{a}$.

The integrand becomes a rational function such as $\dfrac{x^{3}}{x^{4}+1}$.

Step 2: Check the numerator against the denominator's derivative. …

Common Mistakes

Mistake 1: Misreading $e^{3\log x}$.

Why it's wrong: $e^{3\log x}=x^3$, not $3x$ or $3\log x$; a wrong conversion changes the integrand entirely. Correct approach: apply $e^{a\log x}=x^a$.

Mistake 2: Losing the factor $\tfrac14$.

Why it's wrong: since $\tfrac{d}{dx}(x^4+1)=4x^3$, we get $x^3\,dx=\tfrac14\,du$; omitting the $\tfrac14$ overstates the answer by a factor of $4$. Correct approach: carry the constant from $du$. …