Question

Find the integral of $ta{n}^{-1}\left(x\right)$

• A. $xta{n}^{-1}\left(x\right)-xln|1+x^2|+c$
• B. $x^{2}ta{n}^{-1}\left(x\right)-\frac{1}{2}ln|1+x^2|+c$
• C. $xta{n}^{-1}\left(x\right)-\frac{1}{2}ln|1+x^2|+c$
• D. $xta{n}^{-1}\left(x\right)-\frac{x}{2}ln|1+x^2|+c$

Answer 1

The correct option is C $xta{n}^{-1}\left(x\right)-\frac{1}{2}ln|1+x^2|+c$

This is a function, whose integral, we can’t find directly. We saw that in those cases, even if there is only one function, we can apply integration by parts to find the solution by taking ‘1’ as the second function. We are not sure if we will be able to find the integral this way. We will get to know that as we proceed. Another approach to solve this problem would be to proceed by the substitution x = tan(t). We will proceed by applying integration by parts.
$\int ta{n}^{-1}\left(x\right)dx=\int \left(ta{n}^{-1}\right(x)\times 1)dx$
=$ta{n}^1\left(x\right)\int 1dx-\int \left[\frac{d}{dx}\right(ta{n}^1\left(x\right)\int 1dx]dx$
=$ta{n}^{-1}\left(x\right)x-\int \frac{1}{1+x^2}xdx$
=$xta{n}^{-1}\left(x\right)-\frac{1}{2}\int \frac{2x}{1+x^2}dx$
In the second term, if substitute, $t=1+x^2$ , $dt=2xdx$, which is the numerator. So we get
$\int ta{n}^{-1}\left(x\right)dx=xta{n}^{-1}\left(x\right)-\frac{1}{2}\int \frac{dt}{t}$
=$xta{n}^{-1}\left(x\right)-\frac{1}{2}ln|t|$
Substituting back t = 1+$x^2$
$\int ta{n}^{-1}\left(x\right)dx=xta{n}^{-1}\left(x\right)-\frac{1}{2}ln|1+x^2|+c$