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exis [7]
3 years ago
7

Please, I need help in this ??

Mathematics
1 answer:
nignag [31]3 years ago
4 0

Answer:

\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c

Step-by-step explanation:

\int\frac{x^{4}}{x^{4} -1}dx

Adding and Subtracting 1 to the Numerator

\int\frac{x^{4} - 1 + 1}{x^{4} -1}dx

Dividing Numerator seperately by x^{4} - 1

\int 1 + \frac{1}{x^{4}-1 }\, dx

Here integral of 1 is x +c1 (where c1 is constant of integration

x + c1 + \int\frac{1}{(x-1)(x+1)(x^{2}+1)}\, dx----------------------------------(1)

We apply method of partial fractions to perform the integral

\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x^{2} + 1}------------------------------------------(2)

\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)}{(x-1)(x+1)(x^{2} +1)}

1 = A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)-------------------------(3)

Substitute x= 1 , -1 , i in equation (3)

1 = A(1+1)(1+1)

A = \frac{1}{4}

1 = B(-1-1)(1+1)

B = -\frac{1}{4}

1 = C(i-1)(i+1)

C = -\frac{1}{2}

Substituting A, B, C in equation (2)

\int\frac{x^{4}}{x^{4} -1}dx = \int\frac{1}{4(x-1)} - \frac{1}{4(x+1)} -\frac{1}{2(x^{2}+1) }

On integration

Here \int \frac{1}{x}dx = lnx and \int\frac{1}{x^{2}+1 } dx = arctanx

\int\frac{x^{4}}{x^{4} -1}dx = \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2---------------------------------------(4)

Substitute equation (4) back in equation (1) we get

x + c1 + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2

Here c1 + c2 can be added to another and written as c

Therefore,

\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c

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