OTHER CLASS XII TOPICS
Application of Integrals
Finding the area bounded by a curve
Applications of Derivatives
Applications of Derivatives
Approximations
Increasing and Decreasing Functions
Maxima and Minima
Tangents and Normals
Continuity and Differentiability
Chain Rule
Power Rule
Product Rule
Quotient Rule
Integration
Integration by Partial Fractions
Integration of Particular Functions
Integration using Trignometric Identities
Matrices
Diagonal Matrix
Finding the missing number in a Matrix
Identity Matrix
Scalar Matrix
Square Matrix
Probability
Probability
Relations and Functions
Equivalence Relation
Reflexive Relation
Symmetric Relation
Transitive Relation
Three Dimensional Geometry
Finding the direction cosines
Finding the equation of line
Finding the shortest distance
Trigonometry
Inverse Trigonometric Functions
Vector Algebra
Area of parallelogram
Finding the position vector of a point
Finding the unit vector
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Mathematics
Class XII
Integration
Integration of Particular Functions
Question
$\int \dfrac{x^2}{1-x^6}$ $dx$
$\dfrac{1}{3}log \left| \dfrac{1-x^3}{1+x^3} \right|$ + $C$
$\dfrac{1}{6}log \left| \dfrac{1+x^2}{1-x^2} \right|$ + $C$
$\dfrac{1}{6}log \left| \dfrac{1+x^3}{1-x^3} \right|$ + $C$
$\dfrac{1}{6}log \left| \dfrac{1+x^6}{1-x^6} \right|$ + $C$
Validate
Solution
The correct answer is $\dfrac{1}{6}log \left| \dfrac{1+x^3}{1-x^3} \right|$ + $C$
Explanation
$I$ = $\int \dfrac{x^2}{1-x^6}$ $dx$ = $\int \dfrac{x^2}{1-(x^3)^2}$ $dx$
Let us put $x^3=t$
⇒ $3x^2 dx = dt$ or $x^2dx = \dfrac{1}{3}dt$
∴ $I$ = $\int \dfrac{1}{1-t^2} \dfrac{1}{3}dt$
∴ $I$ = $\dfrac{1}{3} \dfrac{1}{2} log \left| \dfrac{1+t}{1-t} \right|$ + $C$
= $\dfrac{1}{6}log \left| \dfrac{1+x^3}{1-x^3} \right|$ + $C$
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