Animated Solution for Mathematics - Indefinite Integration: Evaluate $\int {\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx$

To solve the integral $I=\int {\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx$, we must first simplify the quadratic expression within the denominator.

Consider the expression $4x^2+8x+13$. By factoring out $4$ from the $x$ terms, we obtain $4(x^2+2x)+13$.

Completing the square inside the bracket yields:

Thus, the denominator is transformed into the form:

With the form $X^2+a^2$ identified, where $X=2(x+1)$ and $a=3$, we employ a trigonometric substitution. We set $2(x+1)=3\tan \theta$, which implies $x+1=\frac{3}{2}\tan \theta$.

Geometrically, this corresponds to a right-angled triangle where $\tan \theta =\frac{2x+2}{3}$. The opposite side is $2x+2$, the adjacent side is $3$, and the hypotenuse is $\sqrt{(2x+2{)}^2+3^2}=\sqrt{4x^2+8x+13}$.

This triangle allows us to express the argument of the inverse sine function as:

Next, we differentiate $x+1=\frac{3}{2}\tan \theta$ to find $dx$:

Substituting these into the original integral, the expression ${\sin }^{-1}(\sin \theta )$ simplifies to $\theta$. The integral becomes:

We apply Integration by Parts using the formula $\int udv=uv-\int vdu$. Let $u=\theta$ and $dv={\sec }^2\theta d\theta$.

This gives us:

Recalling that $\int \tan \theta d\theta =\log |\sec \theta |$, the expression becomes:

We now revert to the variable $x$. From our triangle, $\theta ={\tan }^{-1}\left(\frac{2x+2}{3}\right)$ and $\sec \theta =\frac{\sqrt{4x^2+8x+13}}{3}$.

Substituting these back, the first term becomes:

The second term simplifies as follows:

The final result is: