Question:
If $\int\frac{1}{x^{2}+4x+\alpha}dx=\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x+2}{2\sqrt{2}}\right)+c$, then $\int\frac{1}{x^{2}+4x-\alpha}dx=$
If $\int\frac{1}{x^{2}+4x+\alpha}dx=\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x+2}{2\sqrt{2}}\right)+c$, then $\int\frac{1}{x^{2}+4x-\alpha}dx=$
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Use the coefficient from the $\tan^{-1}$ term to quickly find the constant value $a$, then square it to find the missing variable $\alpha$.
Updated On: Jun 3, 2026
- $\frac{1}{4}\log\left|\frac{x-12}{x+2}\right|+k$
- $\frac{1}{8}\log\left|\frac{x-2}{x+6}\right|+k$
- $\frac{1}{8}\log\left|\frac{x+6}{x+8}\right|+k$
- $\frac{1}{4}\log\left|\frac{x-12}{x+16}\right|+k$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
We use the standard integral formula $\int \frac{1}{(x+h)^2 + a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x+h}{a}\right)$.
Step 2: Meaning
Complete the square for the first denominator: $x^2 + 4x + \alpha = (x+2)^2 + (\alpha - 4)$. Comparing the given result $\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x+2}{2\sqrt{2}}\right)$ with our standard formula shows that $a = 2\sqrt{2} \implies a^2 = 8$. Therefore, $\alpha - 4 = 8 \implies \alpha = 12$.
Step 3: Analysis
Now substitute $\alpha = 12$ into the second integral: $\int \frac{1}{x^2 + 4x - 12} dx$. Complete the square for this new denominator: $x^2 + 4x - 12 = (x+2)^2 - 16 = (x+2)^2 - 4^2$. Now apply the standard integral formula $\int \frac{1}{u^2 - a^2} du = \frac{1}{2a}\log\left|\frac{u-a}{u+a}\right|$: $\int \frac{1}{(x+2)^2 - 4^2} dx = \frac{1}{2(4)}\log\left|\frac{(x+2)-4}{(x+2)+4}\right| + k = \frac{1}{8}\log\left|\frac{x-2}{x+6}\right| + k$.
Step 4: Conclusion
This result perfectly matches option (B).
Final Answer: (B)
We use the standard integral formula $\int \frac{1}{(x+h)^2 + a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x+h}{a}\right)$.
Step 2: Meaning
Complete the square for the first denominator: $x^2 + 4x + \alpha = (x+2)^2 + (\alpha - 4)$. Comparing the given result $\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x+2}{2\sqrt{2}}\right)$ with our standard formula shows that $a = 2\sqrt{2} \implies a^2 = 8$. Therefore, $\alpha - 4 = 8 \implies \alpha = 12$.
Step 3: Analysis
Now substitute $\alpha = 12$ into the second integral: $\int \frac{1}{x^2 + 4x - 12} dx$. Complete the square for this new denominator: $x^2 + 4x - 12 = (x+2)^2 - 16 = (x+2)^2 - 4^2$. Now apply the standard integral formula $\int \frac{1}{u^2 - a^2} du = \frac{1}{2a}\log\left|\frac{u-a}{u+a}\right|$: $\int \frac{1}{(x+2)^2 - 4^2} dx = \frac{1}{2(4)}\log\left|\frac{(x+2)-4}{(x+2)+4}\right| + k = \frac{1}{8}\log\left|\frac{x-2}{x+6}\right| + k$.
Step 4: Conclusion
This result perfectly matches option (B).
Final Answer: (B)
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