Complete the square and match a standard integral

Factor out 9: $\dfrac19\int\dfrac1{2/9-x^2}\,dx$. Apply the $a^2-x^2$ logarithmic formula with $a=\sqrt2/3$. The coefficient is $(1/9)/(2a)=1/(6\sqrt2)$.

The integral is $\dfrac1{6\sqrt2}\ln\left|\dfrac{\sqrt2+3x}{\sqrt2-3x}\right|+C$.

Complete the square: $6+4y-y^2=6-[(y-2)^2-4]=10-(y-2)^2$. Apply the difference-of-squares logarithmic form.

The integral is $\dfrac1{2\sqrt{10}}\ln\left|\dfrac{\sqrt{10}+y-2}{\sqrt{10}-y+2}\right|+C$.

The denominator is $(y-2)^2-17$. Use the $(y-2)^2-a^2$ form with $a=\sqrt{17}$, retaining absolute values.

The integral is $\dfrac1{2\sqrt{17}}\ln\left|\dfrac{y-2-\sqrt{17}}{y-2+\sqrt{17}}\right|+C$.

Factor 2 and complete the square: $2y^2-4y-1=2[(y-1)^2-3/2]$. Integrate as $\dfrac12\int[(y-1)^2-3/2]^{-1}\,dy$. The logarithm coefficient is $\dfrac12\cdot\dfrac1{2\sqrt{3/2}}=\sqrt6/12$. Multiply numerator and denominator of its ratio by √2 to get the printed form.

The integral is $\dfrac{\sqrt6}{12}\ln\left|\dfrac{\sqrt2y-\sqrt2-\sqrt3}{\sqrt2y-\sqrt2+\sqrt3}\right|+C$.

Complete the square to $(x+2)^2+16$. Apply the $u^2+a^2$ arctangent form with u=x+2 and a=4.

The integral is $\dfrac14\tan^{-1}\left(\dfrac{x+2}4\right)+C$.

Complete the square: $2x^2+3x+10=2(x+3/4)^2+71/8$. The printed formula box matches $f(x)=\sqrt2(x+3/4)$, $f^\prime(x)=\sqrt2$ and $a^2=71/8$. Multiply and divide the integral by √2 to match $\int f^\prime(x)/(f(x)^2+a^2)\,dx=(1/a)\tan^{-1}(f(x)/a)+C$. Simplification gives the stated coefficient 2/√71 and argument (4x+3)/√71.

The integral is $\dfrac2{\sqrt{71}}\tan^{-1}\left(\dfrac{4x+3}{\sqrt{71}}\right)+C$.

The radicand is $5-[(y+2)^2-4]=9-(y+2)^2$. Apply the inverse-sine form. The original integrand requires −5<y<1.

The integral is $\sin^{-1}\left(\dfrac{y+2}3\right)+C$.

Factor √9 from the denominator to obtain $\dfrac1{\sqrt9}\int[25/9-x^2]^{-1/2}\,dx$. The result is $(1/3)\sin^{-1}(x/(5/3))$, with −5/3<x<5/3.

The integral is $\dfrac13\sin^{-1}\left(\dfrac{3x}5\right)+C$.