Question

integration(x^4+x^2+1/x^2+1)dx​

Answer 1

Answer:

Step-by-step explanation:

I=∫x2+1x4+x2+1dx

Simply pull out x2 common from the numerator as well as the denominator…

I=∫x2(1+1x2)x2(x2+1+1x2)dx

I=∫1+1x2x2+1+1x2dx

Let’s play around with the denominator :D

I=∫1+1x2x2+1+1x2+2−2dx

I=∫1+1x2(x2+1x2−2)+3dx

I=∫1+1x2(x−1x)2+3dx

Now let’s assume t=x−1x

Taking Derivative of Both Sides…

dt=(1+1x2)dx

So, our Integral now becomes…

I=∫1t2+3dt

I=∫1t2+(3–√)2dt

Hey that’s a Standard Integral!!!

I=13–√tan−1t3–√+C

Now all that’s left to do is plug in t=x−1x

I=13–√tan−1(x−1x3–√)+C