Problem to solve
Find the indefinite integral \[\int\frac{x^2}{\sqrt{2x^2-5}} dx\]
Solution
STEP 1
Take the integral: integral x^2/sqrt(2 x^2 - 5) dx
STEP 2
For the integrand x^2/sqrt(2 x^2 - 5), substitute x = sqrt(5/2) sec(u) and dx = sqrt(5/2) tan(u) sec(u) du. Then sqrt(2 x^2 - 5) = sqrt(5 sec^2(u) - 5) = sqrt(5) tan(u) and u = sec^(-1)(sqrt(2/5) x): = sqrt(5/2) integral1/2 sqrt(5) sec^3(u) du
STEP 3
Factor out constants: = 5/(2 sqrt(2)) integral sec^3(u) du
STEP 4
Use the reduction formula, integral sec^m(u) du = (sin(u) sec^(m - 1)(u))/(m - 1) + (m - 2)/(m - 1) integral sec^(-2 + m)(u) du, where m = 3: = (5 tan(u) sec(u))/(4 sqrt(2)) + 5/(4 sqrt(2)) integral sec(u) du
STEP 5
Multiply numerator and denominator of sec(u) by tan(u) + sec(u): = (5 tan(u) sec(u))/(4 sqrt(2)) + 5/(4 sqrt(2)) integral(sec^2(u) + tan(u) sec(u))/(tan(u) + sec(u)) du
STEP 6
For the integrand (sec^2(u) + tan(u) sec(u))/(tan(u) + sec(u)), substitute s = tan(u) + sec(u) and ds = (sec^2(u) + tan(u) sec(u)) du: = (5 tan(u) sec(u))/(4 sqrt(2)) + 5/(4 sqrt(2)) integral1/s ds
STEP 7
The integral of 1/s is log(s): = (5 log(s))/(4 sqrt(2)) + (5 tan(u) sec(u))/(4 sqrt(2)) + constant
STEP 8
Substitute back for s = tan(u) + sec(u): = (5 tan(u) sec(u))/(4 sqrt(2)) + (5 log(tan(u) + sec(u)))/(4 sqrt(2)) + constant
STEP 9
Substitute back for u = sec^(-1)(sqrt(2/5) x): = (5 sec(sec^(-1)(sqrt(2/5) x)) tan(sec^(-1)(sqrt(2/5) x)))/(4 sqrt(2)) + (5 log(sec(sec^(-1)(sqrt(2/5) x)) + tan(sec^(-1)(sqrt(2/5) x))))/(4 sqrt(2)) + constant
STEP 10
Simplify using sec(sec^(-1)(z)) = z and tan(sec^(-1)(z)) = sqrt(1 - 1/z^2) z: = 1/4 sqrt(2 x^2 - 5) x + (5 log((sqrt(2 x^2 - 5) + sqrt(2) x)/sqrt(5)))/(4 sqrt(2)) + constant
STEP 11
Factor the answer a different way: = 1/8 (2 sqrt(2 x^2 - 5) x + 5 sqrt(2) log((sqrt(2 x^2 - 5) + sqrt(2) x)/sqrt(5))) + constant
STEP 12
Which is equivalent for restricted x values to: Answer: | | = 1/8 (2 sqrt(2 x^2 - 5) x + 5 sqrt(2) tanh^(-1)(x/sqrt(x^2 - 5/2))) + constant
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Edited by tododemate® - 2021