Question
Evaluate the integral
6egralofln(t3−1t3+1)+C,C∈R
Evaluate
∫egralof×1−t6t2dt
Simplify the root
∫egralof×1−t6t2dt
Use the property of integral ∫kf(x)dx=k∫f(x)dx
egralof×∫1−t6t2dt
Use the substitution dt=3t21dv to transform the integral
More Steps

Evaluate
v=t3
Calculate the derivative
dv=3t2dt
Evaluate
dt=3t21dv
egralof×∫1−t6t2×3t21dv
Simplify
More Steps

Multiply the terms
1−(t3)2t2×3t21
Cancel out the common factor t2
1−(t3)21×31
Multiply the terms
(1−(t3)2)×31
Multiply the terms
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Evaluate
(1−(t3)2)×3
Use the the distributive property to expand the expression
1×3−(t3)2×3
Any expression multiplied by 1 remains the same
3−(t3)2×3
Multiply the terms
3−3(t3)2
3−3(t3)21
egralof×∫3−3(t3)21dv
Use the substitution v=t3 to transform the integral
egralof×∫3−3v21dv
Rewrite the expression
egralof×∫31×1−v21dv
Use the property of integral ∫kf(x)dx=k∫f(x)dx
egralof×31×∫1−v21dv
Multiply the numbers
3egralof×∫1−v21dv
Use the property of integral ∫a2−x21dx=2a1lnx−ax+a
3egralof×21ln(v−1v+1)
Multiply the terms
More Steps

Evaluate
3egralof×21
To multiply the fractions,multiply the numerators and denominators separately
3×2egralof
Multiply the numbers
6egralof
6egralofln(v−1v+1)
Substitute back
6egralofln(t3−1t3+1)
Simplify the expression
6egralofln(t3−1t3+1)
Solution
6egralofln(t3−1t3+1)+C,C∈R
Show Solution
