1.

Integral X(arctanx)^2 dx

Answer»

∫ sqrt(a^2 + X^2) dx


= ∫ sqrt(a^2 + (au)^2) * a * du


= ∫ a * sqrt(1 + u^2) * a * du


= (a^2) * ∫ sqrt(1 + u^2) du


= 0.5*a^2*[ u*sqrt(1 + u^2) + Ln(u + sqrt(1 + u^2)) ] + c


Now making the reverse substitution u = x/a, we get


0.5*a*x*sqrt(1 + (x/a)^2) + 0.5*a^2*Ln(x/a + sqrt(1 + (x/a)^2)) + c


With a little algebra you can express this in an alternate FORM that is more COMPACT:


[x*sqrt(x^2 + a^2) + a^2*Ln(x + sqrt(x^2 + a^2))]/2 + c


Discussion

No Comment Found

Related InterviewSolutions