Substitution?

Calculus Level 5

$\int \dfrac{\sqrt{1+x^4}}{1-x^4}dx$

If the result of the integral above is in the form:

$a\left(ln(f(x))+ \arcsin(g(x))\right)+c$

(Where a is constant and f,g are functions of x.)

Then the value of

$f(0)+a(g(1))$ is:

1.5 1 1.25 1.75

1 solution

Sid Patak
Jul 10, 2017

$\int \dfrac{\sqrt{1+x^4}}{1-x^4}dx$

$u= \dfrac{\sqrt{2}x}{\sqrt{1+x^4}}$ $du= \dfrac{\sqrt{2}(1-x^4)}{\sqrt{1+x^4}(1+x^4)}dx$

$\dfrac{1}{\sqrt2} \int \dfrac{du}{1-u^4}$

$\dfrac{1}{2\sqrt2} \int \left(\dfrac{1}{1-u^2} + \dfrac{1}{1+u^2}\right)du$

$\dfrac{1}{4\sqrt2} ln\left(\dfrac{1+u}{1-u}\right) + \dfrac{1}{2\sqrt2}arctan(u) +c$

$\dfrac{1}{2\sqrt2}\left(ln\left(\dfrac{\sqrt{1+x^4}+\sqrt2x}{1-x^2}\right) + arcsin\left(\dfrac{\sqrt2x}{1+x^2}\right)\right) +c$

$Required Result = 1 +\dfrac{1}{4} = \dfrac{5}{4}$

$= 1.25$

$u= \dfrac{\sqrt{2}x}{\sqrt{1+x^4}}$

How did you think of this substitution? What motivates you to do so?

Pi Han Goh - 3 years, 11 months ago

How you thought of the substitution ? What made you do that?

Priyanshu Mishra - 3 years, 10 months ago

Interesting. I did the substitution $u=\frac{\sqrt{1+x^4}}{x}$ .

James Wilson - 3 years, 9 months ago