Q.Integrate the following function: $\int (2x - 3\cos x + e^x) dx$

The key idea is that the antiderivative of a sum is the sum of the antiderivatives. We integrate each term separately.

First, $\int 2x \, dx = 2 \cdot \frac{x^2}{2} = x^2$.

Second, $\int -3\cos x \, dx = -3 \sin x$.

Third, $\int e^x \, dx = e^x$. …

The key idea here is that integration is a linear operation. This means that when you have a sum (or difference) of functions inside the integral, you can break it apart and integrate each piece on its own. It’s like unpacking a suitcase — you deal with each item separately instead of trying to lift the whole thing at once.

Let’s look at the three pieces we have: $2x$, $-3\cos x$, and $e^x$. Each one is a standard form whose antiderivative you should know from memory. The only twist is the constants in front — but constants just tag along for the ride.

1. Integrate $2x$. The power rule for integration says: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, provided $n \neq -1$. Here $x$ is $x^1$, so $n=1$.

The constant $2$ cancels neatly with the denominator, leaving just $x^2$. No constant of integration yet — we’ll add one at the very end.

2. Integrate $-3\cos x$. The antiderivative of $\cos x$ is $\sin x$. Why? Because the derivative of $\sin x$ is $\cos x$. So going backwards, $\int \cos x \, dx = \sin x + C$. The constant $-3$ just multiplies the result:

3. Integrate $e^x$. This is the easiest of all. The exponential function $e^x$ is its own derivative and its own antiderivative. So:

No constant factor to worry about here. …

Method: Linearity plus standard antiderivatives

Use this whenever the integrand is a sum or difference of familiar functions (powers, $\sin x$, $\cos x$, $e^x$): integration is linear, so break it apart and integrate each piece with its known formula.

Steps

Step 1: Split the integral across the $+$ and $-$ signs.

and pull any constant coefficient outside its integral.

Step 2: Apply the standard antiderivative to each term. …

Common Mistakes

Mistake 1: Getting the sign of $\int\cos x\,dx$ wrong.

Why it's wrong: $\int\cos x\,dx=\sin x$ (not $-\sin x$), so $\int(-3\cos x)\,dx=-3\sin x$. Confusing it with $\int\sin x\,dx=-\cos x$ flips the sign. Correct approach: memorise the pair — cosine integrates to $+\sin$, sine integrates to $-\cos$.

Mistake 2: Writing three separate constants. …