20.x (r4-1)

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Solution

The integral is given as,

I= ∫ dx x( x 4 −1 )

Multiply numerator and denominator by x 3 .

I= ∫ x 3 dx x 3 ×x( x 4 −1 ) I= ∫ x 3 dx x 4 ( x 4 −1 )

Assume x 4 =t

Differentiate the above with respect to t.

4 x 3 dx=dt x 3 dx= dt 4

Substitute the values in the integral.

I= 1 4 ∫ dt t( t−1 )

Use rule of partial fraction to simplify it.

1 t( t−1 ) = A t + B ( t−1 ) 1=A( t−1 )+Bt

Substitute t=0then,

A=−1

Substitute t=1then,

B=1

Substitute the values and integrate,

I= 1 4 ∫ −dt t + 1 4 ∫ dt ( t−1 ) I= 1 4 [ −log| t |+log| t−1 | ]+C

By substituting x 4 for t, we get

I= 1 4 [ −log| x 4 |+log| x 4 −1 | ]+C I= 1 4 log| x 4 −1 x 4 |+C

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