Partial fractions and derivative matching

Factor the denominator as $(x-3)(x+1)$ and write $A/(x-3)+B/(x+1)$. Then $x=A(x+1)+B(x-3)$. At x=3, A=3/4; at x=−1, B=1/4. Integrate these two simple fractions.

The integral is $\dfrac34\ln|x-3|+\dfrac14\ln|x+1|+C$.

Write $A/y+B/y^2+C/(y-2)$. Clearing denominators gives $1=Ay(y-2)+B(y-2)+Cy^2$. The source uses y=0,2,1 to obtain B=−1/2, C=1/4, A=−1/4. Thus integrate $-1/(4y)-1/(2y^2)+1/[4(y-2)]$. This gives $-\ln|y|/4+1/(2y)+\ln|y-2|/4$, which combines as stated.

The integral is $\dfrac1{2y}+\dfrac14\ln\left|\dfrac{y-2}y\right|+C$.

Write $(Ax+B)/(x^2+x+1)+C/(x-2)$. From $1=(Ax+B)(x-2)+C(x^2+x+1)$, inputs x=2,0,1 give C=1/7, B=−3/7, A=−1/7. The integrand becomes $-(x+3)/[7(x^2+x+1)]+1/[7(x-2)]$. Split $x+3=(2x+1)/2+5/2$ and complete the square $x^2+x+1=(x+1/2)^2+3/4$. Integrate the derivative term as a logarithm, the remainder as arctangent and the last simple pole as ln|x−2|. The quadratic is positive for all real x, so its logarithm needs no absolute value.

The integral is $-\dfrac1{14}\ln(x^2+x+1)-\dfrac5{7\sqrt3}\tan^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)+\dfrac17\ln|x-2|+C$.

Rewrite $16x+5=2(8x+2)+1$. The derivative part integrates to $2\ln(4x^2+2x+1)$. For the remainder use $4x^2+2x+1=4[(x+1/4)^2+3/16]$; the arctangent coefficient is (1/4)/(√3/4)=1/√3.

The integral is $2\ln(4x^2+2x+1)+\dfrac1{\sqrt3}\tan^{-1}\left(\dfrac{4x+1}{\sqrt3}\right)+C$.

Split $s+4=(4s+2)/4+7/2$. The first term integrates to a quarter logarithm. For the second, $2s^2+2s+3=2[(s+1/2)^2+5/4]$. The remaining coefficient 7/4 multiplied by 2/√5 gives 7/(2√5) in the arctangent term.

The integral is $\dfrac14\ln(2s^2+2s+3)+\dfrac7{2\sqrt5}\tan^{-1}\left(\dfrac{2s+1}{\sqrt5}\right)+C$.

Write $2x-1=(2x+4)-5$. Integrate the derivative-over-denominator term as a logarithm, and complete the square $x^2+4x+5=(x+2)^2+1$ for the remaining arctangent.

The integral is $\ln(x^2+4x+5)-5\tan^{-1}(x+2)+C$.