Question

The value of the integral \(\int \frac{x^{2}+1}{(x+1)^{2}}dx=\) ............

• A. \(x-2\,\log |x+1|-\frac{2}{x+1}+c\)
• B. \(x-2\,\log |x+1|+\frac{2}{x+1}+c\)
• C. \(x+2\,\log |x+1|+c\)
• D. \(x-2\,\log |x+1|+c\)

Hint:

Write the numerator as (x + 1) squared minus 2x, then split into partial fractions.

Answer 1

The Correct Option is A

Step 1: Substitute t = x + 1
Then $x=t-1$ and $x^{2}+1=t^{2}-2t+2$.

Step 2: Split the fraction
$\frac{t^{2}-2t+2}{t^{2}}=1-\frac{2}{t}+\frac{2}{t^{2}}$.

Step 3: Integrate and go back
$\int=t-2\log|t|-\frac{2}{t}+c=x+1-2\log|x+1|-\frac{2}{x+1}+c$.
The constant 1 is absorbed in $c$.

Final Answer:
Hence the integral is x - 2 log|x + 1| - 2/(x + 1) + c. \[ \boxed{\text{(A) }x-2\,\log |x+1|-\dfrac{2}{x+1}+c} \]