evaluate each of the following integral:-
I= ∫((x^2+ 1)/(x^4+ 1))dx
Y=∫(sinhxcoshx)dx/√(sinhx^4+coshx^4)
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To evaluate the integral I= ∫((x^2+ 1)/(x^4+ 1))dx, we can use partial fraction decomposition and then integrate term by term.
1. Begin by factoring the denominator, x^4 + 1, using the difference of squares formula: x^4 + 1 = (x^2 + 1)(x^2 - 1).
2. Rewrite the integral as a sum of two fractions:
I = ∫((x^2+ 1)/(x^4+ 1))dx = ∫(A/(x^2 + 1))dx + ∫(B/(x^2 - 1))dx, where A and B are constants to be determined.
3. Find the values of A and B by expanding the numerator on the right-hand side of the equation and comparing like terms with the original numerator:
x^2 + 1 = A(x^2 - 1) + B(x^2 + 1)
4. Simplify the equation by collecting terms with x^2:
x^2 + 1 = (A + B)x^2 + (B - A)
5. Equate the coefficients of x^2 and the constant terms on both sides to obtain two equations:
A + B = 1 (coefficient of x^2 terms)
B - A = 1 (constant term)
6. Solve the system of equations to find the values of A and B. In this case, adding the two equations together eliminates the A term:
2B = 2, which gives B = 1. Substituting this value into the second equation gives A = 0.
7. Rewrite the integral using the partial fraction decomposition:
I = ∫(0/(x^2 + 1))dx + ∫(1/(x^2 - 1))dx
8. Integrate each term separately:
∫(0/(x^2 + 1))dx = 0 (integral of a constant is zero)
∫(1/(x^2 - 1))dx = arctan(x) + C (integral of 1/(x^2 - 1) is arctan(x))
9. Finally, the solution is:
I = 0 + arctan(x) + C
To evaluate the integral Y = ∫(sinh(x)cosh(x))/(√(sinh(x)^4 + cosh(x)^4))dx, we can use a trigonometric substitution.
1. Let u = sinh(x) and du = cosh(x)dx. Rewrite the integral in terms of u:
Y = ∫(u/cosh(x))/√(u^4 + cosh(x)^4) du
2. Simplify the denominator of the integrand using the identity cosh(x)^2 - sinh(x)^2 = 1:
Y = ∫(u/cosh(x))/√(u^4 + (cosh(x)^2 - sinh(x)^2)^2) du = ∫(u/cosh(x))/√(u^4 + cosh(x)^4 - 2sinh(x)^2cosh(x)^2) du
3. Substitute du = cosh(x)dx and simplify the integral:
Y = ∫(u/cosh(x))/√(u^4 + u^2 - 2u^2) du = ∫(u/cosh(x))/√(u^2(u^2 + 1 - 2)) du = ∫(u/cosh(x))/√(u^2(u^2 - 1)) du
4. Factor out u^2 from the square root:
Y = ∫u/(u√(u^2 - 1)) du = ∫1/√(u^2 - 1) du
5. Notice that the integrand corresponds to the derivative of the inverse hyperbolic sine function, as the derivative of arcsinh(u) is 1/√(u^2 + 1). Therefore, we can rewrite the integral in terms of arcsinh(u):
Y = arcsinh(u) + C
6. Substituting back u = sinh(x) gives the final result:
Y = arcsinh(sinh(x)) + C = x + C
Therefore, the solution to the integral Y is x + C, where C is the constant of integration.