Q.(a) Evaluate: $\displaystyle\int\dfrac{3x+5}{x^2+4x+7}\,dx$ OR

$\displaystyle\int\dfrac{3x+5}{x^2+4x+7}\,dx=\dfrac32\ln(x^2+4x+7)-\dfrac{1}{\sqrt3}\tan^{-1}\!\left(\dfrac{x+2}{\sqrt3}\right)+C$.

The derivative of the denominator is $2x+4$. Write the numerator as a multiple of this plus a constant:

$3x+5=\dfrac32(2x+4)+B\ \Rightarrow\ 3x+5=3x+6+B\ \Rightarrow\ B=-1.$

So

$\int\dfrac{3x+5}{x^2+4x+7}\,dx=\dfrac32\int\dfrac{2x+4}{x^2+4x+7}\,dx-\int\dfrac{1}{x^2+4x+7}\,dx.$

First integral (numerator is the derivative of the denominator):

$\dfrac32\int\dfrac{2x+4}{x^2+4x+7}\,dx=\dfrac32\ln(x^2+4x+7)+C_1.$

Second integral: complete the square, $x^2+4x+7=(x+2)^2+3$: …