Question

If $\int \frac{3x+2}{4x^2+4x+5}dx = A\log(4x^2+4x+5)+B\tan^{-1}(\frac{2x+1}{2})+c$, then $A+B=$

• A. 1/2
• B. 3/4
• C. 3/8
• D. 1/8

Hint:

To integrate rational functions of the form $\int \frac{px+q}{ax^2+bx+c}dx$, always start by expressing the numerator as a linear combination of the derivative of the denominator and a constant: $px+q = \lambda(2ax+b) + \mu$. This splits the problem into a logarithm part and an inverse tangent part.

Answer 1

The Correct Option is A

Step 1: Express the numerator in terms of the derivative of the denominator.
The integral is of the form $\int \frac{px+q}{ax^2+bx+c}dx$. Let the numerator be $3x+2 = \lambda \frac{d}{dx}(4x^2+4x+5) + \mu$. The derivative of the denominator is $8x+4$. \[ 3x+2 = \lambda(8x+4) + \mu = 8\lambda x + (4\lambda + \mu). \] Equating coefficients of $x$: $8\lambda=3 \implies \lambda = 3/8$. Equating constant terms: $4\lambda+\mu=2 \implies 4(3/8)+\mu=2 \implies 3/2+\mu=2 \implies \mu = 1/2$.

Step 2: Split the integral into two parts.
The integral can now be written as: \[ I = \int \frac{\frac{3}{8}(8x+4) + \frac{1}{2}}{4x^2+4x+5}dx = \frac{3}{8}\int \frac{8x+4}{4x^2+4x+5}dx + \frac{1}{2}\int \frac{1}{4x^2+4x+5}dx. \]

Step 3: Evaluate the first integral.
The first part is of the form $\int \frac{f'(x)}{f(x)}dx = \log|f(x)|$. \[ \frac{3}{8}\int \frac{8x+4}{4x^2+4x+5}dx = \frac{3}{8}\log(4x^2+4x+5). \] By comparing with the given form, we find $A = 3/8$.

Step 4: Evaluate the second integral.
The second part is $I_2 = \frac{1}{2}\int \frac{1}{4x^2+4x+5}dx$. We complete the square in the denominator. \[ 4x^2+4x+5 = (2x)^2+2(2x)(1)+1^2 + 4 = (2x+1)^2+4. \] \[ I_2 = \frac{1}{2}\int \frac{1}{(2x+1)^2+2^2}dx. \] Let $u=2x+1$, then $du=2dx \implies dx=du/2$. \[ I_2 = \frac{1}{2} \int \frac{1}{u^2+2^2} \frac{du}{2} = \frac{1}{4} \int \frac{1}{u^2+2^2}du. \] This is a standard integral of the form $\int \frac{1}{t^2+a^2}dt = \frac{1}{a}\tan^{-1}(\frac{t}{a})$. \[ I_2 = \frac{1}{4} \left[ \frac{1}{2}\tan^{-1}\left(\frac{u}{2}\right) \right] = \frac{1}{8}\tan^{-1}\left(\frac{2x+1}{2}\right). \] By comparing with the given form, we find $B=1/8$.

Step 5: Calculate A+B.
\[ A+B = \frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}. \] \[ \boxed{A+B = \frac{1}{2}}. \]