Q.Integrate the following function: $\frac{1}{x(x^n + 1)}$ [Hint: multiply numerator and denominator by $x^{n-1}$ and put $x^n = t$]

Concept: U Substitution – the hint suggests a clever manipulation to create a derivative inside the integral.

Step 1: Multiply numerator and denominator by $x^{n-1}$:

Step 2: Let $t = x^n$, so $dt = n x^{n-1} dx$, giving $x^{n-1} dx = \frac{dt}{n}$. The integral becomes:

Step 3: Use partial fractions: $\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$. Integrate: …

Why this approach works

When you see a function like $\frac{1}{x(x^n+1)}$, the denominator is a product of $x$ and a binomial in $x^n$. Direct substitution of $x^n = t$ is tempting, but $dx$ doesn't play nicely with $dt$ unless we adjust the integrand first. The hint — multiply numerator and denominator by $x^{n-1}$ — is the key move. Why $x^{n-1}$? Because $x^{n-1} dx$ is exactly $\frac{1}{n} dt$ when $t = x^n$. That turns the integral into something rational in $t$, which we can split using partial fractions.

Let's walk through it.

1. Multiply numerator and denominator by $x^{n-1}$

   We start with:

Multiply top and bottom by $x^{n-1}$:

The denominator is now $x^n (x^n + 1)$ — a product of two factors, each a power of $x^n$. That's the signal: substitute $t = x^n$.

2. Perform the substitution $x^n = t$

   Differentiate: $n x^{n-1} dx = dt$, so $x^{n-1} dx = \frac{1}{n} dt$. The integral becomes:

Clean and simple.

3. Decompose into partial fractions

   We need to split $\frac{1}{t(t+1)}$. Write:

Multiply through by $t(t+1)$:

Solve for $A$ and $B$. Set $t = 0$: $1 = A(1) \Rightarrow A = 1$. Set $t = -1$: $1 = B(-1) \Rightarrow B = -1$. So:

$\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$ …

Method: Manufacture a Substitution, then Partial Fractions

Use this when an integrand isn't directly rational but becomes rational after a clever substitution — often signalled by a hint like "put $x^n=t$."

Steps

Step 1: Create the derivative you need.

Multiply numerator and denominator by a factor that makes the derivative of the intended substitution appear. For $\dfrac{1}{x(x^n+1)}$, multiply by $x^{n-1}$:

Step 2: Substitute.

Let $t=x^n$, so $dt=n\,x^{n-1}dx$, i.e. $x^{n-1}dx=\dfrac{dt}{n}$. The integral becomes $\dfrac1n\displaystyle\int\frac{dt}{t(t+1)}$. …

Common Mistakes

Mistake 1: Dropping the $\tfrac1n$ factor from the substitution.

Why it's wrong: $t=x^n$ gives $dt=n\,x^{n-1}dx$, so $x^{n-1}dx=\dfrac{dt}{n}$ — the $\tfrac1n$ multiplies the whole integral; omitting it scales the answer wrongly. Correct approach: carry $\dfrac1n$ through to $\dfrac1n\log\left|\dfrac{x^n}{x^n+1}\right|$.

Mistake 2: Substituting $t=x^n$ without first multiplying by $x^{n-1}$. …