How to Solve These Two Indefinite Integrals?

[juantheron]

$(a)\;\;:: \displaystyle \int\frac{1}{\left(x+\sqrt{x\cdot (x+1)}\right)^2}dx$

$(b)\;\;::\displaystyle \int\frac{1}{(x^4-1)^2}dx$

My Trial :: (a) $\displaystyle \int\frac{1}{(x+\sqrt{x\cdot (x+1)})^2}dx$

$\displaystyle \int\frac{1}{x\left(\sqrt{x}+\sqrt{x+1}\right)^2}dx = \int\frac{1}{x\cdot (x+x+1+2\sqrt{x^2+x})}dx$

$\displaystyle = \frac{1}{2}\int\frac{1}{x\cdot \left(x+0.5 + \sqrt{(x+0.5)^2-(0.5)^2}\right)}dx$

Now I did not understand how can i solve it,

Help me

Thanks

Similarly for (b) $\displaystyle \int\frac{1}{(x^2+1)^2\cdot (x+1)^2 \cdot (x-1)^2}$

But Using Partial fraction, It become very Complex, Is any other way by which we cal solve it

please explain here

Thanks

 

[Chris L T521]

Re: 2 Indefinite Integrals

$(b)\;\;::\displaystyle \int\frac{1}{(x^4-1)^2}dx$

Similarly for (b) $\displaystyle \int\frac{1}{(x^2+1)^2\cdot (x+1)^2 \cdot (x-1)^2}$

But Using Partial fraction, It become very Complex, Is any other way by which we cal solve it

please explain here

Thanks

Here's a way to rewrite the integrand. It seems to have the flavor of decomposing the fraction by partial fractions, but it's still kinda long. I'm doing it this way because I don't feel like solving a system of equations involving 8 unknown constants. XD

Note that

$\displaystyle \begin{aligned}\frac{1}{(x^4-1)^2} &= \frac{1}{4}\left[\frac{((x^2+1)-(x^2-1))^2}{(x^2+1)^2(x^2-1)^2}\right] \\ &= \frac{1}{4}\left[\frac{(x^2+1)^2 -2(x^2+1)(x^2-1) + (x^2-1)^2}{(x^2+1)^2(x^2-1)^2}\right]\\ &= \frac{1}{4}\left[ \frac{1}{(x^2-1)^2} - \frac{2}{(x^2+1)(x^2-1)} + \frac{1}{(x^2+1)^2}\right] \\ &= \frac{1}{4}\left[ \frac{1}{4}\frac{((x+1)-(x-1))^2}{(x+1)^2(x-1)^2} - \frac{(x^2+1)-(x^2-1)}{(x^2+1)(x^2-1)} + \frac{1}{(x^2+1)^2}\right]\\ &= \frac{1}{4}\left[\frac{1}{4}\left(\frac{1}{(x-1)^2}- \frac{2}{(x+1)(x-1)} + \frac{1}{(x+1)^2}\right) - \frac{1}{x^2-1} + \frac{1}{x^2+1} + \frac{1}{(x^2+1)^2}\right] \\ &= \frac{1}{4}\left[\frac{1}{4}\left(\frac{1}{(x-1)^2}- \frac{1}{x-1} + \frac{1}{x+1} + \frac{1}{(x+1)^2}\right) - \frac{1}{2}\frac{1}{x-1} + \frac{1}{2}\frac{1}{x+1} + \frac{1}{x^2+1} + \frac{1}{(x^2+1)^2}\right] \\ &= \frac{1}{16(x-1)^2} - \frac{3}{16(x-1)} + \frac{3}{16(x+1)} +\frac{1}{16(x+1)^2} + \frac{1}{4(x^2+1)} + \frac{1}{4(x^2+1)^2}\end{aligned}$

The integrals of each of these fractions should be rather straightforward. However, you'll need to apply a trig substitution when you evaluate $\displaystyle\int \frac{1}{4(x^2+1)^2}\,dx$.

I hope this makes sense and makes things (a little) easier! (Bigsmile)

 

[Prove It]

$(a)\;\;:: \displaystyle \int\frac{1}{\left(x+\sqrt{x\cdot (x+1)}\right)^2}dx$

$(b)\;\;::\displaystyle \int\frac{1}{(x^4-1)^2}dx$

My Trial :: (a) $\displaystyle \int\frac{1}{(x+\sqrt{x\cdot (x+1)})^2}dx$

$\displaystyle \int\frac{1}{x\left(\sqrt{x}+\sqrt{x+1}\right)^2}dx = \int\frac{1}{x\cdot (x+x+1+2\sqrt{x^2+x})}dx$

$\displaystyle = \frac{1}{2}\int\frac{1}{x\cdot \left(x+0.5 + \sqrt{(x+0.5)^2-(0.5)^2}\right)}dx$

Now I did not understand how can i solve it,

Help me

Thanks

Similarly for (b) $\displaystyle \int\frac{1}{(x^2+1)^2\cdot (x+1)^2 \cdot (x-1)^2}$

But Using Partial fraction, It become very Complex, Is any other way by which we cal solve it

please explain here

Thanks

$$\displaystyle \begin{align*} \int{ \frac{1}{ \left[ x + \sqrt{ x \left( x + 1 \right) } \right] ^2}\,dx } &= \int{ \frac{ \left[ x - \sqrt{x\left( x + 1 \right) } \right] ^2 }{ \left[ x + \sqrt{x \left( x + 1 \right) } \right] ^2 \left[ x - \sqrt{ x \left( x + 1 \right) } \right] ^2 } } \\ &= \int{ \frac{\left[ x - \sqrt{x \left( x + 1 \right) } \right] ^2}{ \left[ x^2 - x \left( x + 1 \right) \right] ^2 } \, dx} \\ &= \int{ \frac{x^2 - 2x\sqrt{x \left( x + 1 \right) } + x \left( x + 1 \right) }{ \left( x^2 - x^2 - x \right) ^2 } \, dx } \\ &= \int{ \frac{x^2 -2x \sqrt{ x \left( x + 1 \right) } + x^2 + x }{ \left( -x \right) ^2 } \,dx } \\ &= \int{ \frac{x\left[ 1 - 2 \sqrt{ x \left( x + 1 \right) } \right] }{ x^2 } \,dx } \\ &= \int{ \frac{1 - 2\sqrt{x \left( x + 1 \right) }}{x}\,dx } \\ &= \int{ \frac{1}{x}\,dx} - \int{ 2 \frac{\sqrt{ x + 1}}{\sqrt{x}} \,dx } \end{align*}$$

Can you go from here?