Integration by u substitution doubts: ∫ 1 / (2x2 + 1) dx

[pencile]

∫ 1 / (2x² + 1) dx


since ∫ (1 / u² + 1) du = tan^-1 u + c
let 2x² = u²
u = sqrt2 x
du = sqrt2 dx


(1/sqrt2 * du = numerator * dx)


∫ (1 / 2x² + 1) dx
= 1/sqrt2 tan^-1 (sqrt2 x) + c


In another question, however


∫ 1 / ((x + 1/6)² + 11/36) dx
= ∫ 1 / [11/36 (36/11 (x + 1/6)² + 1)] dx


since ∫ 1 / (u² + 1) du = tan^-1 u + c
let 36/11 (x + 1/6)² = u²
u = 6x + 1 / sqrt11
du = 6/sqrt11 dx


36/11 ∫ 1 / [(36/11 (x + 1/6)² + 1)] dx


(11*sqrt11/36*6 * 36/11 * du = numerator * dx)


= 11*sqrt11/36*6 tan^-1 (6x + 1 / sqrt11) + c


Correct answer = 6/sqrt11 tan^-1 (6x + 1 / sqrt11) + c


What am I missing? Thanks for helping.

 

[tkhunny]

For starters, you should realize that 1 / x^2 + 1 = \(\displaystyle \dfrac{1}{x^{2}}+1\), and is likely not what you had in mind.

Second, use your actual definition to find the differential relationship. You should be aware that x = u is NOT the same as x^2 = u^2.