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The integration by parts formula is as follow-

$$ \int u\ dv = uv- \int v\ du$$

normally, I use L.I.A.T.E. As a guide for the priority of the choice of $u$

Where L is Log, I is inverse trigo, A is algebra, T is trigo, E is exponential.

The left side will be $u$ and right side will be $v$

Now, the question is $\int \tan^{-1} x\ dx$

What’s the product of 2 function here ?

In that question, I only see an inverse trigo function. $x$ cannot be counted as an algebra because x is part of the inverse trigo function.

so how do I pick $u$ ?

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  • $\begingroup$ $u=\tan^{-1}$ and $dv=dx$ do the trick $\endgroup$ – marwalix Apr 22 '18 at 8:26
  • $\begingroup$ @Ma rwalix so where does ‘u’ lie on L.I.A.T.E. ? $\endgroup$ – user59439 Apr 22 '18 at 8:27
  • $\begingroup$ I edited my comment $\endgroup$ – marwalix Apr 22 '18 at 8:28
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    $\begingroup$ @user59439 Since $x^0 = 1$, I guess it's an algebraic function. $\endgroup$ – Toby Mak Apr 22 '18 at 8:31
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Take $u=\tan^{-1}(x)$ and $dv=dx$

$$du={dx\over x^2+1}$$

$$\int\tan^{-1}(x)dx=x\tan^{-1}(x)-\int{xdx \over x^2+1}$$

Now in the integral in the RHS substitute $w=x^2+1$ so $dw=2xdx$...

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I.L.A.T.E = Inverse trig . Logarithmic . Algebraic. Trig. Exponential

Here you select $u=\arctan(x) $ and $dv= 1$

$I=\int\arctan(x)\,dx$

$I = x\arctan(x)-\int\frac{x}{x^2+1}\,dx$

To evaluate $J = \int\frac x{x^2+1}\,dx$

let $x^2+1 = u\implies du = 2x\,dx$

$J = \frac12\int\frac 1u\,du$

$ J = \frac12\ln(x^2+1)+C$

$I =x\arctan(x) -\frac12\ln(x^2+1)+C$

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  • $\begingroup$ It should be $I(x)=x\arctan{x}-{1\over 2}\log(x^2+1)+c$ $\endgroup$ – marwalix Apr 22 '18 at 8:40
  • $\begingroup$ oops sorry , will edit at once. Thank you $\endgroup$ – The Integrator Apr 22 '18 at 8:42

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