Q.Integrate the following function: $\tan^{-1}x$

The key idea is to treat $\tan^{-1}x$ as $1 \cdot \tan^{-1}x$ and apply integration by parts.

Let $u = \tan^{-1}x$ and $dv = dx$. Then $du = \frac{1}{1+x^2}\,dx$ and $v = x$.

Using $\int u\,dv = uv - \int v\,du$:

For the remaining integral, substitute $t = 1+x^2$, so $dt = 2x\,dx$ and $x\,dx = \frac{dt}{2}$: …

Why Integration by Parts?

The function $\tan^{-1}x$ (also written as $\arctan x$) is an inverse trigonometric function. There is no direct formula for its integral — you cannot reverse-differentiate it by inspection. But integration by parts gives us a way: it lets us trade a hard integral for an easier one.

The rule is:

The trick is to choose $u$ and $dv$ so that the new integral $\int v \, du$ is simpler. For $\tan^{-1}x$, the derivative $\frac{d}{dx} \tan^{-1}x = \frac{1}{1+x^2}$ is a rational function — much easier to integrate than the original inverse trig function. So we set:

• $u = \tan^{-1}x$ (so $du = \frac{1}{1+x^2} \, dx$)
• $dv = 1 \cdot dx$ (so $v = x$)

This is the classic choice: the inverse trig function becomes the $u$, and the "1" becomes the $dv$.

Step-by-step solution

1. Set up integration by parts We have:

Choose:

Then:

2. Apply the formula

Substituting:

So:

3. Solve the new integral …

Method: Integration by parts on a single inverse-trig (or log) function

When the integrand is a lone inverse-trig or logarithmic function ($\tan^{-1}x$, $\sin^{-1}x$, $\log x$, ...) with nothing multiplying it, treat it as that function times $1$ and integrate by parts.

Steps

Step 1: Insert the hidden factor of $1$.

Step 2: Assign the parts.

Take the given function as $u$ (so you differentiate it) and $1$ as $dv$:

Step 3: Apply $\int u\,dv=uv-\int v\,du$. …

Common Mistakes

Mistake 1: Believing by parts needs two visible factors, so giving up.

Why it's wrong: any single inverse-trig/log function is integrable by treating it as $f(x)\cdot 1$. Correct approach: set $dv=dx$, $v=x$.

Mistake 2: Setting $u=1,\ dv=\tan^{-1}x\,dx$.

Why it's wrong: this asks you to integrate $\tan^{-1}x$ — the very unknown. Correct approach: differentiate the inverse-trig function, integrate the $1$. …