Integration by parts: "math is broken"

Question

just trying to solve a small example on integration by parts, and a weird thing happens: I come to an original expression increased by one. Please help me find out where the flaw is!

The task is to calculate the following indefinite integral: $$ \int\tan^{-1}x\text{d}x $$

Integration by parts formula (just in case): $$ \int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)\text{d}x $$

Let's expand our original integral: $$ \int\tan^{-1}x\text{d}x = \int\cos x \sin^{-1}x\text{d}x $$

If $$ f(x) = \sin^{-1}x $$ $$ g'(x) = \cos x $$ then $$ f'(x) = -\sin^{-2}x\cos x $$ $$ g(x) = \sin x $$

Applying integration by parts formula: $$ \int\cos x \sin^{-1}x\text{d}x = \sin^{-1}x\sin x - \int-\sin^{-2}x\cos x\sin x\text{d}x = 1 + \int\tan^{-1}x\text{d}x $$

So, where have I made a mistake?

Answer 1

You haven't made a mistake.

Remember how integrals have " ${}+C$ " at the end? This is why. The ${}+C$ 'absorbs' all constants together into an unknown constant. Indefinite integrals don't give you a single function: they give you a set of functions that differ by a constant.

Answer 2

So, $\tan^{-1}(x)$ is not equal to $\dfrac{1}{\tan(x)}$. Instead, that is $\cot(x)$.

For the integration by parts, let $u=\tan^{-1}(x)$ and $dv=1\quad\!\!\!\!dx$

Remember that the formula is $uv-\int vdu$