Question

integration of x^4+1/x^6+1

Answer 1

I don't know about it

Answer 2

The given function is…

f(x)=x4+1x6+1f(x)=x4+1x6+1

=(x2+1)2–2x2(x2)3+13=(x2+1)2–2x2(x2)3+13

=(x2+1)2–2x2(x2+1)(x4−x2+1)=(x2+1)2–2x2(x2+1)(x4−x2+1)

=x2+1x4–x2+1−2x2(x3)2+1=x2+1x4–x2+1−2x2(x3)2+1

⇒f(x)=1+1x2x2+1x2−1−2x2(x3)2+1⇒f(x)=1+1x2x2+1x2−1−2x2(x3)2+1

Now differentiating both sides w.r.t x,we get…

∫f(x)dx=∫(1+1x2)dx(x−1x)2+2x⋅1x−1−2∫x2dx1+(x3)2∫f(x)dx=∫(1+1x2)dx(x−1x)2+2x⋅1x−1−2∫x2dx1+(x3)2

=∫d[x−1x](x−1x)2+12−23∫d(x3)1+(x3)2=∫d[x−1x](x−1x)2+12−23∫d(x3)1+(x3)2

=tan−1(x−1x)−23tan−1(x3)+C=tan−1⁡(x−1x)−23tan−1⁡(x3)+C



I=∫(x2+1)2−2x2(x2+1)(x4−x2+1)dxI=∫(x2+1)2−2x2(x2+1)(x4−x2+1)dx

I=∫x2+1x4−x2+1dx−2∫x2(x3)2+1dxI=∫x2+1x4−x2+1dx−2∫x2(x3)2+1dx

I=∫1x2+1(x−1x)2+1dx−2∫x2(x3)2+1dxI=∫1x2+1(x−1x)2+1dx−2∫x2(x3)2+1dx

Now let x−1x=tx−1x=t

Taking Derivative of Both Sides…

(1+1x2)dx=dt(1+1x2)dx=dt

Also make a second substitution x3=ux3=u

Once again taking derivative of both sides…

3x2dx=du3x2dx=du

Let’s take a look at the new face of our Integral…

I=∫1t2+1dt−23∫1u2+1duI=∫1t2+1dt−23∫1u2+1du

I=tan−1t−23tan−1u+CI=tan−1⁡t−23tan−1⁡u+C

Now put back the values of tt and uu

I=tan−1(x−1x)−23tan−1x3+CI=tan−1⁡(x−1x)−23tan−1⁡x3+C

And there we have it!