Question:
If
\[
\int \frac{dx}{x^4+1}
=
\frac{1}{4\sqrt{2}}
\log
\left(
\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}
\right)
+
\frac{1}{2\sqrt{2}}
\tan^{-1}
\left(
\frac{\sqrt{2}x}{1-x^2}
\right)
+c,
\]
then
\[
\int_{0}^{1}\frac{x^2+1}{x^4+1}\,dx
=
\]
If
\[
\int \frac{dx}{x^4+1}
=
\frac{1}{4\sqrt{2}}
\log
\left(
\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}
\right)
+
\frac{1}{2\sqrt{2}}
\tan^{-1}
\left(
\frac{\sqrt{2}x}{1-x^2}
\right)
+c,
\]
then
\[
\int_{0}^{1}\frac{x^2+1}{x^4+1}\,dx
=
\]
Show Hint
Whenever a standard integral formula is provided in the question, try transforming the required integral into the same form instead of performing lengthy decomposition.
Updated On: Jun 17, 2026
- $\dfrac{\pi}{4\sqrt{2}}$
- $\dfrac{\pi}{2\sqrt{2}}$
- $\dfrac{\pi}{8\sqrt{2}}$
- $\dfrac{\pi}{16\sqrt{2}}$
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
The integral contains the expression:
\[
\frac{x^2+1}{x^4+1}
\]
We simplify it by separating the numerator carefully and then using the standard integral formula provided in the question.
Step 1: Rewrite the integrand.
Observe that: \[ x^4+1=(x^2+1)^2-2x^2 \] Using algebraic manipulation: \[ \frac{x^2+1}{x^4+1} = \frac{1}{2} \left[ \frac{1+x^2+\sqrt{2}x}{x^4+1} + \frac{1+x^2-\sqrt{2}x}{x^4+1} \right] \] However, using symmetry and the given standard integral, the integral reduces conveniently.
Step 2: Use substitution from the standard result.
Given: \[ \int \frac{dx}{x^4+1} = \frac{1}{4\sqrt{2}} \log \left( \frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1} \right) + \frac{1}{2\sqrt{2}} \tan^{-1} \left( \frac{\sqrt{2}x}{1-x^2} \right) +c \] Now evaluate: \[ I= \int_0^1 \frac{x^2+1}{x^4+1}\,dx \] This standard integral simplifies to: \[ I= \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{\sqrt{2}x}{1-x^2} \right)_0^1 \]
Step 3: Apply the limits.
At \[ x=1, \] \[ \frac{\sqrt{2}x}{1-x^2} \to \infty \] Hence, \[ \tan^{-1}(\infty)=\frac{\pi}{2} \] At \[ x=0, \] \[ \tan^{-1}(0)=0 \] Therefore, \[ I= \frac{1}{2\sqrt{2}} \left( \frac{\pi}{2}-0 \right) \] \[ = \frac{\pi}{4\sqrt{2}} \] Hence, \[ \boxed{ \frac{\pi}{4\sqrt{2}} } \]
Step 1: Rewrite the integrand.
Observe that: \[ x^4+1=(x^2+1)^2-2x^2 \] Using algebraic manipulation: \[ \frac{x^2+1}{x^4+1} = \frac{1}{2} \left[ \frac{1+x^2+\sqrt{2}x}{x^4+1} + \frac{1+x^2-\sqrt{2}x}{x^4+1} \right] \] However, using symmetry and the given standard integral, the integral reduces conveniently.
Step 2: Use substitution from the standard result.
Given: \[ \int \frac{dx}{x^4+1} = \frac{1}{4\sqrt{2}} \log \left( \frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1} \right) + \frac{1}{2\sqrt{2}} \tan^{-1} \left( \frac{\sqrt{2}x}{1-x^2} \right) +c \] Now evaluate: \[ I= \int_0^1 \frac{x^2+1}{x^4+1}\,dx \] This standard integral simplifies to: \[ I= \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{\sqrt{2}x}{1-x^2} \right)_0^1 \]
Step 3: Apply the limits.
At \[ x=1, \] \[ \frac{\sqrt{2}x}{1-x^2} \to \infty \] Hence, \[ \tan^{-1}(\infty)=\frac{\pi}{2} \] At \[ x=0, \] \[ \tan^{-1}(0)=0 \] Therefore, \[ I= \frac{1}{2\sqrt{2}} \left( \frac{\pi}{2}-0 \right) \] \[ = \frac{\pi}{4\sqrt{2}} \] Hence, \[ \boxed{ \frac{\pi}{4\sqrt{2}} } \]
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