Question
Answers
# int 1/(sqrt(2x^2+5)) \ dx = 1/sqrt(2) \ sinh^(-1)(sqrt(2)/sqrt(5)x) + C # Explanation:
We have:
# I = int 1/(sqrt(2x^2+5)) \ dx # We can simplify and standardise the quadratic in the denominator.
Let
#u=sqrt(2)/sqrt(5)x => (du)/dx = sqrt(2)/sqrt(5) # Substituting into the integral we get:
# I = int 1/(sqrt(2(sqrt(5)/sqrt(2)u)^2+5) )\ (sqrt(5)/sqrt(2)) \ du #
# \ \ = sqrt(5)/sqrt(2) \ int 1/( sqrt(5u^2+5) ) \ du #
# \ \ = sqrt(5)/sqrt(2) \ int 1/( sqrt(5)sqrt(u^2+1) ) \ du #
# \ \ = 1/sqrt(2) \ int 1/( sqrt(u^2+1) ) \ du # This is now a standard integral, so we have:
# I = 1/sqrt(2) \ sinh^(-1)u + C # And, restoring the substitution, we get:
# I = 1/sqrt(2) \ sinh^(-1)(sqrt(2)/sqrt(5)x) + C #
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