Q.Integrate the following function: $\frac{(x-3)e^x}{(x-1)^3}$

Step 1 — Simplify the exponential

Step 2 — Combine with $2^x$ …

Concept & Intuition

At first glance, the integrand looks messy — a sum of sine and cosine over a square root containing a double-angle identity. The key is to notice that $1 + 2\sin x \cos x = \sin^2 x + \cos^2 x + 2\sin x \cos x = (\sin x + \cos x)^2$. That square under the square root suggests cancellation, but we must be careful: $\sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x|$, not simply $\sin x + \cos x$. However, the problem likely assumes the principal branch where the expression simplifies nicely, and the multiple-choice options hint that the answer is a simple function.

Step-by-step solution

1. Simplify the denominator Recall the Pythagorean identity: $\sin^2 x + \cos^2 x = 1$. Also, $2\sin x \cos x = \sin 2x$. Then

So the integrand becomes

2. Handle the square root For real $x$, $\sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x|$. Therefore the integrand is

This equals $1$ when $\sin x + \cos x > 0$, and $-1$ when $\sin x + \cos x < 0$.

Over intervals where the sign is constant, the integral is $\pm x + C$.

3. Choose the principal branch In typical indefinite integration problems (especially multiple-choice), we assume the expression is taken on an interval where $\sin x + \cos x > 0$ (e.g., $-\pi/4 < x < 3\pi/4$), so that the square root cancels directly to $\sin x + \cos x$. Then $$$$

Write $x^x=e^{x\log x}$. Then

$\frac{d}{dx}\,x^x=e^{x\log x}\cdot\frac{d}{dx}(x\log x)=x^x(\log x+1).$ …

Step 1 — Convert the numerator using the double-angle identity.

With $\cos 2\theta = 2\cos^{2}\theta - 1$:

(Note $\alpha$ is a constant, so $\cos 2\alpha$ and $\cos\alpha$ are constants throughout.)

Step 2 — Factor as a difference of squares.

Step 3 — Cancel with the denominator.

This is the whole trick: what looked like a hard quotient is really a linear expression in $\cos x$.

Step 4 — Integrate. …

The key insight here is that $\sin 2x$ is not a basic function of $\sin x$ alone — but it is $2\sin x \cos x$, and $\cos x$ is exactly the derivative of $\sin x$. That suggests a substitution: let $t = \sin x$, so $dt = \cos x \, dx$. The integral then becomes something you can handle with integration by parts.

Let’s walk through it.

1. Rewrite the integrand $\sin 2x = 2\sin x \cos x$, so

2. Substitute $t = \sin x$ Then $dt = \cos x \, dx$, and $\sin x = t$. The integral becomes

3. Integrate by parts For $\int t e^{t} \, dt$, let $u = t$, $dv = e^{t} dt$. Then $du = dt$, $v = e^{t}$.

4. Multiply by 2 and substitute back

Replace $t$ with $\sin x$:

$\int e^{\sin x} \sin 2x \, dx = 2 e^{\sin x}(\sin x - 1) + C.$ …