solve-x-4-x-2-1-2-x-2-1-dx-

Question Number 187855 by Michaelfaraday last updated on 23/Feb/23

s o l v e

\int \frac{x^{4} + x^{2} + 1}{2 (x^{2} + 1)} d x

Answered by cortano12 last updated on 23/Feb/23

I = \frac{1}{2} \int \frac{{(x^{2} + 1)}^{2} - x^{2}}{x^{2} + 1} dx

I = \frac{1}{6} x^{3} + \frac{1}{2} x - \frac{1}{2} \int \frac{x^{2}}{x^{2} + 1} dx

= \frac{1}{6} x^{3} + \frac{1}{2} x - \frac{1}{2} \int \frac{\tan^{2} θ}{\sec^{2} θ} \sec^{2} θ d θ; x = \tan θ

= \frac{1}{6} x^{3} + \frac{1}{2} x - \frac{1}{2} \int (\sec^{2} θ - 1) d θ

= \frac{1}{6} x^{3} + \frac{1}{2} x - \frac{1}{2} x + \frac{1}{2} \arctan x + C

= \frac{1}{6} x^{3} + \frac{1}{2} \arctan x + C

Commented by Michaelfaraday last updated on 25/Feb/23

t h a n k s s i r

Answered by CElcedricjunior last updated on 23/Feb/23

\int \frac{x^{4} + x^{2} + 1}{2 (x^{2} + 1)} d x = \frac{1}{2} \int [\frac{x^{2} (x^{2} + 1)}{(x^{2} + 1)} + \frac{1}{(x^{2} + 1)}] d x

= \frac{1}{2} \int [x^{2} + \frac{1}{x^{2} + 1}] d x M o i v r e

= \frac{1}{2} [\frac{1}{3} x^{3} + a r c t a n (x)] + k ★ C e d r i c j u n i o r

Commented by Michaelfaraday last updated on 25/Feb/23

t h a n k s s i r