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Anonymous Thu 19 Dec 12:41:16 2019 No.11238004 View Reply Original Report

Quoted By: >>11238010 >>11238105

Anyone bigbrain pls help

Anonymous

Anonymous Thu 19 Dec 2019 12:43:07 No.11238008 Report

Quoted By:

1+x-1 = x
apply linearity.

Anonymous

Anonymous Thu 19 Dec 2019 12:44:24 No.11238010 Report

Quoted By:

>>11238004
easy

$-2\left(\dfrac{\ln\left(\left|\sqrt{\frac{x+1}{x}}-1\right|\right)-\ln\left(\sqrt{\frac{x+1}{x}}+1\right)}{16}+\dfrac{-3\sqrt{\frac{x+1}{x}}+\frac{x+1}{x}-2}{8\left(\frac{x+1}{x}\right)^\frac{3}{2}-8\sqrt{\frac{x+1}{x}}+\frac{8\left(x+1\right)}{x}-8}\right)$

Anonymous

Anonymous Thu 19 Dec 2019 12:44:26 No.11238011 Report

Quoted By:

Exercise from my first semester midterm.

Anonymous

Anonymous Thu 19 Dec 2019 13:57:28 No.11238105 Report

Quoted By: >>11238121 >>11238146 >>11238150

>>11238004
This is just a basic ass euler substitution.

$\frac{x+1}{x} = t^{2}$

$\int \frac{t}{1+t}\frac{-2t}{(t^{2}-1)^{2}}dt$

Anonymous

Anonymous Thu 19 Dec 2019 14:12:28 No.11238121 Report

Quoted By:

>>11238105
From there, just factorize the denominator into
$(t+1)^{3}$ and $(t-1)^{2}$ , apply integration by parts, and then use iterated polynomial division to express the numerator as a linear combination of $(t+1)^{k}$ , or $(t-1)^{k}$ polynomials (with real coefficients). Then, cancel the numerator terms with terms in the denominator, so, the integral then becomes a sum of $\frac{a_{k}}{(t+1)^{k}}$ , and $\frac{b_{k}}{(t-1)^{k}}$ , which can be easily solved.

Anonymous

Anonymous Thu 19 Dec 2019 14:28:26 No.11238137 Report

Quoted By:

is that really how you draw your 1's, what is wrong with you?

Anonymous

Anonymous Thu 19 Dec 2019 14:34:28 No.11238146 Report

Quoted By:

>>11238105
To solve this, factorize the denominator into $(t+1)^{3}$ and $(t-1)^{2}$ , then, use the extended Euclidian algorithm for polynomials to find polynomials p(t) and q(t) such that $p(t)(t-1)^{2} + q(t)(t+1)^{3} = 1$ , multiply both sides by $-2t^{2}[\math], giving you <script type="math/tex">-2t^{2}p(t)(t-1)^{2} -2t^{2}q(t)(t+1)^{3} = -2t^{2}$ . Substitute this into the numerator, and cancel the factors in the numerator, with factors in the denominator. Then, use iterated polynomial division on $-2t^{2}p(t)[\math] and <script type="math/tex">-2t^{2}q(t)$ to express them as linear combinations of $(t+1)^{k}$ and $(t-1)^{k}$ , with real numbers as coefficients. Finally, cancel those polynomials in the numerator, with factors in the denominator, giving you some easy integrals for you to solve.

Anonymous

Anonymous Thu 19 Dec 2019 14:37:06 No.11238150 Report

Quoted By:

>>11238105
To solve this, factorize the denominator into $(t+1)^{3}$ and $(t-1)^{2}$ , then, use the extended Euclidian algorithm for polynomials to find polynomials p(t) and q(t) such that $p(t)(t-1)^{2} + q(t)(t+1)^{3} = 1$ , multiply both sides by $-2t^{2}$ , giving you $-2t^{2}p(t)(t-1)^{2} -2t^{2}q(t)(t+1)^{3} = -2t^{2}$ . Substitute this into the numerator, and cancel the factors in the numerator, with factors in the denominator. Then, use iterated polynomial division on $-2t^{2}p(t)$ and $-2t^{2}q(t)$ to express them as linear combinations of $(t+1)^{k}$ and $(t-1)^{k}$ , with real numbers as coefficients. Finally, cancel those polynomials in the numerator, with factors in the denominator, giving you some easy integrals for you to solve.

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