Solve the integral of y=x^3/(x^4+1)^5 .

### 2 Answers | Add Yours

We have to find the integral of y = x^3/(x^4+1)^5.

Use substitution to find the integral.

Int [ x^3/(x^4+1)^5 dx]

let x^4 + 1 = y

=> 4*x^3 dx = dy

=> x^3 dx = (1/4) dy

=> (1/4)*Int [ (1/y^5) dy]

=> (1/4)*y^(-4)/(-4)

=> (-1/16)/y^4 + C

substitute y = x^4 + 1

=> (-1/16)/(x^4 + 1)^4 + C

**The required integral is (-1/16)/(x^4 + 1)^4 + C**

integral of y=x^3/(x^4+1)^5

We'll solve the integral using substitution technique. We'll note x^4 + 1 = t.

We'll differentiate both sides:

4x^3dx = dt

We'll put:

x^3dx = dt/4

We'll write the integral in t:

Int dt/4t^5 = (1/4)Int t^-5dt

(1/4)Int t^-5dt = (1/4)*t^(-5+1)/(-5+1) +C

Int dt/4t^5 = -(1/16*t^4) + C

We'll substitute t = x^4 + 1

**Int x^3*dx/(x^4+1)^5 = -1/16*(x^4 + 1)^4 + C**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes