Q.Integrate the following function: $\sqrt{x^2+4x+6}$

Complete the square: $4x-x^2=4-(x-2)^2$, then use $\int\frac{dx}{\sqrt{a^2-u^2}}=\sin^{-1}\frac{u}{a}$.

Complete the square under the root: $4x-9x^2=9\big[(\tfrac29)^2-(x-\tfrac29)^2\big]$.

Complete the square under the root and use $\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\frac{u}{a}$; result $\frac12\sin^{-1}\frac{8x-9}{9}$.

Complete the square inside the square root and reduce to the standard form $\int dx/\sqrt{x^2-a^2}$.

$5x^2-2x=5\left[\left(x-\dfrac15\right)^2-\dfrac{1}{25}\right]$, so

$\displaystyle\int\frac{dx}{\sqrt{5x^2-2x}}=\frac{1}{\sqrt5}\int\frac{dx}{\sqrt{(x-1/5)^2-(1/5)^2}}=\frac{1}{\sqrt5}\ln\left|\left(x-\frac15\right)+\sqrt{(x-1/5)^2-1/25}\right|+C$

Multiplying inside by 5 (absorbing the constant into $C$): $=\dfrac{1}{\sqrt5}\log\left|(5x-1)+\sqrt{25x^2-10x}\right|+C$.

Complete the square in the denominator to get the standard $\frac{1}{x^2+a^2}$ form.

$x^2+2x+5=(x+1)^2+4$.

$\int\frac{dx}{(x+1)^2+2^2}=\frac{1}{2}\tan^{-1}\left(\frac{x+1}{2}\right)+C$.

Complete the square under the root and use $\int \tfrac{dx}{\sqrt{t^2-a^2}} = \log|t+\sqrt{t^2-a^2}|$.

$(x-1)(x-2) = x^2-3x+2 = \left(x-\tfrac32\right)^2 - \tfrac14.$

With $t = x-\tfrac32$:

$\int \dfrac{dx}{\sqrt{t^2-(1/2)^2}} = \log\left|t + \sqrt{t^2-\tfrac14}\right| = \log\left|\left(x-\tfrac32\right)+\sqrt{x^2-3x+2}\right|.$

Complete the square and use $\int\sqrt{t^2-a^2}\,dt = \tfrac{t}{2}\sqrt{t^2-a^2} - \tfrac{a^2}{2}\log|t+\sqrt{t^2-a^2}|$.

$x^2+4x+1 = (x+2)^2 - 3$, so $t = x+2$, $a^2 = 3$.

$\int\sqrt{t^2-3}\,dt = \dfrac{t}{2}\sqrt{t^2-3} - \dfrac{3}{2}\log|t+\sqrt{t^2-3}|.$

Substituting back:

$= \dfrac{x+2}{2}\sqrt{x^2+4x+1} - \dfrac{3}{2}\log\left|x+2+\sqrt{x^2+4x+1}\right|.$

Complete the square under the root to get the standard form $\int\dfrac{dx}{\sqrt{a^2-u^2}}=\sin^{-1}\dfrac{u}{a}+C$.

$2x-x^2=-(x^2-2x)=-(x^2-2x+1-1)=1-(x-1)^2$.

$\displaystyle\int\dfrac{dx}{\sqrt{2x-x^2}}=\int\dfrac{dx}{\sqrt{1-(x-1)^2}}=\sin^{-1}(x-1)+C$.

Apply the standard formula $\int\sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$.

Here $a=4$, so $\frac{a^2}{2}=8$. Substituting: $\int\sqrt{16-x^2}\,dx = \frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\frac{x}{4}+C$.