Question

solve this qn ... class 12....please, need urgent ...no wrong answer # thanks​

Answer 1

Given ∫ [1/(x^4+ 1)] dx

The denominator is a sum of squares, thus it is not factorable, therefore partial fraction method seems inapplicable; nevertheless, you can apply it on condition that you rearrange the integrand as follows:

Add and subtract 2x² at denominator, in order to complete

the square:

∫ [1/ (x^4+ 2x² +1 - 2x²)] dx =

∫ {1/ [(x^4+ 2x² +1) - 2x²]} dx =

∫ {1/ [(x² +1)² - 2x²]} dx =

Now factor the denominator, viewing it as a difference

Between two squares:

∫ {1/ [(x² +1)² - (√2x)²]} dx =

∫ {1/ {[(x² +1) + (√2x)] [(x² +1) - (√2x)]}} dx =

∫ {1/ [(x² + √2x +1)(x² - √2x +1)]} dx =

Well, this rearranged integrand is solvable by partial fractions:

So you have to find four real numbers A, B, C and D, so that:

1/ [(x² +√2x +1)(x² -√2x +1)] = (Ax + B) /(x² +√2x +1) + (Cx + D) /(x² -√2x +1) →

1/ [(x² +√2x +1)(x² -√2x +1)] =

[(Ax + B)(x² -√2x +1) + (Cx + D)(x² +√2x +1)] / [(x² +√2x +1)(x² -√2x +1)] →

Equating the numerators:

1 = [(Ax + B)(x² -√2x +1) + (Cx + D)(x² +√2x +1)] →

1 = Ax³ -√2Ax² + Ax + Bx² -√2Bx + B + Cx³+ √2Cx²+ Cx + Dx² +√2Dx + D →

1 = (A + C)x³ + (-√2A + B + √2C + D)x²+ (A -√2B + C +√2D)x + (B + D) →

leading to the system:

| A + C = 0 → A = - C → A = 1/(2√2)

| -√2A + B +√2C+ D = 0 → -√2(- C)+ B +√2C+ D = 0 → 2√2C + 1 = 0 → C = -1/(2√2)

| A -√2B + C +√2D = 0 → - C -√2B + C +√2D = 0 → √2D = √2B → D = B → B = (1/2)

| B + D = 1 → D + D = 1 → D = (1/2)

Now, plugging in the found values, you get (see above):

1/ [(x² +√2x +1)(x² -√2x +1)] = (Ax + B) /(x² +√2x +1) + (Cx + D) /(x² -√2x +1) →

1/ [(x² +√2x +1)(x² -√2x +1)] =

{[1/(2√2)]x + (1/2)} /(x² +√2x +1) + {[-1/(2√2)]x + (1/2)} /(x² -√2x +1) →

Factoring out [1/(2√2)] at both fractions:

1/ [(x² +√2x +1)(x² -√2x +1)] =

[1/(2√2)] [(x +√2) /(x² +√2x +1)] + [1/(2√2)] [(-x +√2) /(x² -√2x +1)] →

Therefore:

∫ {1/ [(x² + √2x +1)(x² - √2x +1)]} dx =

∫ {[1/(2√2)] [(x +√2) /(x² +√2x +1)] + [1/(2√2)] [(-x +√2) /(x² -√2x +1)]} dx =

Breaking it up and taking the constants out:

[1/(2√2)] ∫ [(x +√2) /(x² +√2x +1)] dx + [1/(2√2)] ∫ [(-x +√2)} /(x² -√2x +1)] dx =

Now, in order to change each numerator into the derivative of the respective denominator, divide and multiply the first integral by 2 and the second one by (-2), yielding:

[1/(2√2)](1/2) ∫ [2(x+√2) /(x²+√2x+1)] dx+ [1/(2√2)](-1/2) ∫ [(-2)(-x+√2)} /(x²-√2x+1)] dx =

[1/(4√2)] ∫ [(2x + 2√2) /(x²+√2x +1)] dx - [1/(4√2)] ∫ [(2x - 2√2)} /(x²-√2x +1)] dx =

Then rewrite the numerators as:

[1/(4√2)] ∫ [(2x +√2 +√2) /(x²+√2x +1)] dx - [1/(4√2)] ∫ [(2x - √2 - √2)} /(x²-√2x +1)] dx =

And distribute them as:

[1/(4√2)] ∫ {[(2x +√2)/(x²+√2x +1)] + [√2 /(x²+√2x +1)]} dx -

[1/(4√2)] ∫ {[(2x - √2)/(x²-√2x +1)] - [√2 /(x²-√2x +1)]} dx =

Breaking it up,

[1/(4√2)] ∫ [(2x +√2)/(x²+√2x +1)] dx + [1/(4√2)] ∫ [√2 /(x²+√2x +1)] dx -

[1/(4√2)] ∫ [(2x - √2)/(x²-√2x +1)] dx + [1/(4√2)] ∫ [√2 /(x²-√2x +1)] dx =

Simplifying;

[1/(4√2)] ∫ [d(x²+√2x +1)] /(x²+√2x +1) + (1/4) ∫ [1 /(x²+√2x +1)] dx -

[1/(4√2)] ∫ [d(x²-√2x +1)] /(x²-√2x +1)] + (1/4) ∫ [1 /(x²-√2x +1)] dx =

[1/(4√2)] ln (x²+√2x +1) + (1/4) ∫ [1 /(x²+√2x +1)] dx - [1/(4√2)] ln (x²-√2x +1) +

(1/4) ∫ [1 /(x²-√2x +1)] dx =

[1/(4√2)] [ln (x²+√2x +1) - ln (x²-√2x +1)] + (1/4) ∫ [1 /(x²+√2x +1)] dx +

(1/4) ∫ [1 /(x²-√2x +1)] dx =

According to log properties,

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + (1/4) ∫ [1/(x²+√2x +1)] dx +

(1/4) ∫ [1/(x²-√2x +1)] dx =

To solve the remaining integrals, you need to complete the squares at denominators, since they aren't factorable; more exactly, you have to attempt to change the denominators into {[f (x)]²+ 1} form (which is then integrable as arctan: ∫ {d[f(x)]} /{[f(x)]² + 1} = arctan[f(x)] + c);

Thus, first, multiply and divide the remaining integrals by 2:

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + (1/4)(2) ∫ dx/[2(x²+√2x +1)] +

(1/4)(2) ∫ dx/ [2(x²-√2x +1)] =

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + (1/2) ∫ dx/(2x²+2√2x +2) +

(1/2) ∫ dx/ (2x²-2√2x +2) =

(rewriting 2 as 1+1)

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + (1/2) ∫ dx/(2x² + 2√2x +1 +1) +

(1/2) ∫ dx/ (2x²- 2√2x +1 + 1) =

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + (1/2) ∫ dx/[(2x² + 2√2x +1) +1] +

(1/2) ∫ dx/ [(2x²- 2√2x +1) +1] =

Having completed the squares,

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + (1/2) ∫ dx/[(√2x +1)²+1] +

(1/2) ∫ dx/ [(√2x -1)²+1] =

Finally, divide and multiply the integrals by √2:

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + (1/2)(1/√2) ∫ (√2dx)/[(√2x +1)² +1] +

(1/2)(1/√2) ∫ (√2dx)/ [(√2x -1)² +1] =

So that you can rewrite the numerators as:

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + [1/(2√2)] ∫ [d(√2x +1)] /[(√2x +1)² +1] +

[1/(2√2)] ∫ [d(√2x - 1)] / [(√2x -1)² +1] =

[1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] + [1/(2√2)] arctan (√2x +1) +

[1/(2√2)] arctan (√2x - 1) + c

In conclusion,

∫ [1/(x^4+ 1)] dx = [1/(4√2)] ln [(x²+√2x +1)/(x²-√2x +1)] +

[1/(2√2)] arctan (√2x +1) + [1/(2√2)] arctan (√2x - 1) + c "

Answer 2

Answer:

dear above answer is correct..

Step-by-step explanation:

have a great day dued...♡