Q.Find the following integrals:

Concept: Power Rule Integration — $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$, plus standard forms for $e^x$ and $1/x$.

(i) $\int \frac{x^3 - 1}{x^2}\, dx = \int \left( x - x^{-2} \right) dx$

Integrate termwise: $\frac{x^2}{2} - \frac{x^{-1}}{-1} + C = \frac{x^2}{2} + \frac{1}{x} + C$.

(ii) $\int \left( x^{2/3} + 1 \right) dx = \frac{x^{5/3}}{5/3} + x + C = \frac{3}{5}x^{5/3} + x + C$.

(iii) $\int \left( x^{3/2} + 2e^x - \frac{1}{x} \right) dx = \frac{x^{5/2}}{5/2} + 2e^x - \log|x| + C = \frac{2}{5}x^{5/2} + 2e^x - \log|x| + C$.

The core idea here is the Power Rule for Integration: for any real number $n \neq -1$,

When $n = -1$, the rule gives $\int \frac{1}{x} \, dx = \log|x| + C$. Also, the integral of $e^x$ is itself: $\int e^x \, dx = e^x + C$.

The trick is to first rewrite each integrand so that every term is in the form $x^n$ (or a standard function like $e^x$ or $1/x$). Then integrate term by term, and combine the constants into a single $C$.

Let’s go through each part.

(i) $\int \dfrac{x^3 - 1}{x^2}\, dx$

1. Rewrite the fraction. Split the numerator over the denominator:

Now each term is a simple power of $x$.

2. Integrate term by term.

   • For $x^1$: $\int x \, dx = \frac{x^{2}}{2} + C_1$.
   • For $x^{-2}$: $\int x^{-2} \, dx = \frac{x^{-1}}{-1} = -\frac{1}{x} + C_2$. But note the minus sign in front: we have $-\int x^{-2} dx = -(-\frac{1}{x}) = +\frac{1}{x}$.

   So:

(ii) $\int \left(x^{2/3} + 1\right) dx$

1. Recognise the terms.

   The integrand is already a sum: $x^{2/3}$ and the constant $1$ (which is $x^0$).

2. Apply the Power Rule.

   • For $x^{2/3}$: $n = \frac{2}{3}$, so $n+1 = \frac{5}{3}$.

• For $1$: $\int 1 \, dx = x + C_2$.

Combine:

(iii) $\int \left(x^{3/2} + 2e^x - \dfrac{1}{x}\right) dx$

1. Identify each term’s rule.

   • $x^{3/2}$: power rule with $n = \frac{3}{2}$.
   • $2e^x$: constant times $e^x$, integral is $2e^x$.
   • $-\frac{1}{x}$: this is $-x^{-1}$, so use the special case $n = -1$: $\int \frac{1}{x} dx = \log|x|$.

2. Integrate.

   • $\int x^{3/2} \, dx = \frac{x^{5/2}}{5/2} = \frac{2}{5} x^{5/2} + C_1$.
   • $\int 2e^x \, dx = 2e^x + C_2$.
   • $\int -\frac{1}{x} \, dx = -\log|x| + C_3$.

   Putting it together:

Method: Term-by-Term Power Rule After Rewriting

Use this for integrals of algebraic sums, quotients, or roots: first turn every term into a plain power $x^n$ (or a recognised standard form), then integrate each term separately.

Steps

Step 1: Rewrite the integrand as a sum of powers.

Split fractions and convert roots to fractional exponents. For instance $\dfrac{x^3-1}{x^2}=x-x^{-2}$, and $\sqrt[3]{x^2}=x^{2/3}$. The integral of a sum is the sum of the integrals.

Step 2: Apply the power rule to each term.

Remember the special standard forms that are not powers: $\int e^x\,dx=e^x$ and $\int\frac1x\,dx=\log|x|$.

Step 3: Combine and add one constant.

Add the integrated terms and collect every arbitrary constant into a single $C$ at the end.

Common Mistakes

Mistake 1: Integrating $\dfrac{1}{x}$ with the power rule as $\dfrac{x^{0}}{0}$.

Why it's wrong: the power rule is undefined at $n=-1$. Correct approach: $\int\frac1x\,dx=\log|x|+C$.

Mistake 2: Mishandling the sign when integrating $-x^{-2}$.

Why it's wrong: $\int -x^{-2}\,dx=-\frac{x^{-1}}{-1}=+\frac1x$; students often drop the double-negative and write $-\frac1x$. Correct approach: track the sign carefully — the result is $+\dfrac1x$.

Mistake 3: Trying to "integrate the quotient directly" without splitting.

Why it's wrong: there is no quotient rule for integration, so $\int\frac{f}{g}\neq\frac{\int f}{\int g}$. Correct approach: simplify the quotient into separate power terms first, then integrate.