Question
Evaluate the integral
Diverges
Evaluate
∫3131−x28dx
Use function domain and discontinuity points to transform the expression with the formula ∫acf(x)dx=∫abf(x)dx+∫bcf(x)dx
∫3111−x28dx+∫131−x28dx
By definition,rewrite the improper integral using one-sided limit and a definite integral
a→1−lim(∫31a1−x28dx)+∫131−x28dx
By definition,rewrite the improper integral using one-sided limit and a definite integral
a→1−lim(∫31a1−x28dx)+a→1+lim(∫a31−x28dx)
Evaluate the integral
More Steps

Evaluate
∫31a1−x28dx
Evaluate the integral
∫1−x28dx
Rewrite the expression
∫8×1−x21dx
Use the property of integral ∫kf(x)dx=k∫f(x)dx
8×∫1−x21dx
Use the property of integral ∫a2−x21dx=2a1lnx−ax+a
8×21ln(x−1x+1)
Multiply the terms
More Steps

Evaluate
8×21
Reduce the numbers
4×1
Simplify
4
4ln(x−1x+1)
Return the limits
(4ln(x−1x+1))31a
Substitute the values into formula
4ln(a−1a+1)−4ln31−131+1
Add the numbers
More Steps

Evaluate
31+1
Calculate
33+1
Reduce fractions to a common denominator
33+33
Write all numerators above the common denominator
33+3
4ln(a−1a+1)−4ln31−133+3
Subtract the numbers
More Steps

Evaluate
31−1
Calculate
33−1
Reduce fractions to a common denominator
33−33
Write all numerators above the common denominator
33−3
4ln(a−1a+1)−4ln33−333+3
Calculate the absolute value
More Steps

Calculate
33−333+3
Simplify
3−33+3
Since 3−33+3<0,the absolute value of 3−33+3 is 3−33+3
3−33+3
4ln(a−1a+1)−4ln(3−33+3)
Rewrite the expression
ln(a−1a+1)×4−ln(3−33+3)×4
Factor the expression
(ln(a−1a+1)−ln(3−33+3))×4
Subtract the terms
More Steps

Evaluate
ln(a−1a+1)−ln(3−33+3)
Use logax−logay=logayx to transform the expression
ln3−33+3a−1a+1
Divide the terms
ln(3+3)∣a−1∣(3−3)∣a+1∣
ln(3+3)∣a−1∣(3−3)∣a+1∣×4
Multiply the terms
4ln(3+3)∣a−1∣(3−3)∣a+1∣
a→1−lim4ln(3+3)∣a−1∣(3−3)∣a+1∣+a→1+lim(∫a31−x28dx)
Evaluate the integral
More Steps

Evaluate
∫a31−x28dx
Evaluate the integral
∫1−x28dx
Rewrite the expression
∫8×1−x21dx
Use the property of integral ∫kf(x)dx=k∫f(x)dx
8×∫1−x21dx
Use the property of integral ∫a2−x21dx=2a1lnx−ax+a
8×21ln(x−1x+1)
Multiply the terms
More Steps

Evaluate
8×21
Reduce the numbers
4×1
Simplify
4
4ln(x−1x+1)
Return the limits
(4ln(x−1x+1))a3
Substitute the values into formula
4ln(3−13+1)−4ln(a−1a+1)
When the expression in absolute value bars is not negative, remove the bars
4ln(3−13+1)−4ln(a−1a+1)
Rewrite the expression
ln(3−13+1)×4−ln(a−1a+1)×4
Factor the expression
(ln(3−13+1)−ln(a−1a+1))×4
Subtract the terms
More Steps

Evaluate
ln(3−13+1)−ln(a−1a+1)
Use logax−logay=logayx to transform the expression
lna−1a+13−13+1
Divide the terms
ln(3−1)∣a+1∣(3+1)∣a−1∣
ln(3−1)∣a+1∣(3+1)∣a−1∣×4
Multiply the terms
4ln(3−1)∣a+1∣(3+1)∣a−1∣
a→1−lim4ln(3+3)∣a−1∣(3−3)∣a+1∣+a→1+lim4ln(3−1)∣a+1∣(3+1)∣a−1∣
Evaluate the limit
More Steps

Evaluate
a→1−lim4ln(3+3)∣a−1∣(3−3)∣a+1∣
Rewrite the expression
4×a→1−limln(3+3)∣a−1∣(3−3)∣a+1∣
Calculate
More Steps

Evaluate
a→1−limln(3+3)∣a−1∣(3−3)∣a+1∣
Rewrite the expression
lna→1−lim(3+3)∣a−1∣(3−3)∣a+1∣
Calculate
ln(+∞)
Calculate
+∞
4(+∞)
Simplify
+∞
(+∞)+a→1+lim4ln(3−1)∣a+1∣(3+1)∣a−1∣
Evaluate the limit
More Steps

Evaluate
a→1+lim4ln(3−1)∣a+1∣(3+1)∣a−1∣
Rewrite the expression
4×a→1+limln(3−1)∣a+1∣(3+1)∣a−1∣
Calculate
More Steps

Evaluate
a→1+limln(3−1)∣a+1∣(3+1)∣a−1∣
Rewrite the expression
lna→1+lim(3−1)∣a+1∣(3+1)∣a−1∣
Calculate
ln(0)
Calculate
−∞
4(−∞)
Simplify
−∞
(+∞)+(−∞)
Solution
Diverges
Show Solution
