Integrating $\frac{0.36h^2 + 1.44h + 1.44}{0.034 - 0.012136 \sqrt{h}}$ - Stuck with one part

Question

I have been trying to follow how to integrate $\frac{0.36h^2 + 1.44h + 1.44}{0.034 - 0.012136 \sqrt{h}}$ using

https://www.integral-calculator.com/ ,

but when it gets to

$(2 \div 5295931061521 \times 1517^4)$ Integral $(u + 4250) ​[2301289 u^4 + 39121913000 u^3 + 4 (1517^4 + 62350548843750) u^2 + 34000 (1517^4 + 20783516281250) u + 81455156 * 1517^4 + 750804525660156250000 ] \frac{du} {u}$**

I cannot see how they get $81455156 \times 1517^4$. I can get all the other terms, but not this one.

These are my calculations below, to get to this point

Going from the part of Solving Integral $\frac{(h^2 + 4h +4 ) dh}{1517 * \sqrt{h} - 4250}$

Now we want to find u for the substitution

substitute $u = (1517 \times \sqrt h - 4250 )$ so, as a substitute $u = (1517 \times \sqrt h - 4250 )$

so, $\frac{du}{dh} = ( 1517 \times \frac{du}{dh} [\sqrt h] ) + \frac{du}{dh} [ - 4250 ] = 1517 \times 1/2 h^{1/2 - 1} + 0$

Therefore $\frac{du}{dh} = \frac{(1517 \times {1/2} )} {\sqrt h}$, Multiply du / dh by 2 top and bottom, so, du / dh = ( 1517 * 2 * 1/2 ) / ( 2 * sqrt h ) = 1517 / ( 2 * sqrt h ) dh / du = reciprocal of 1517 / ( 2 * sqrt h ) so dh / du = ( 2 * sqrt h ) / 1517

so, dh = 2 * sqrt h * du / 1517

Use u = (1517 * sqrt h - 4250) from before and add 4250 to both sides to get rid of it on right hand side

u + 4250 = (1517 * sqrt h - 4250 + 4250 ) , so u + 4250 = ( 1517 * sqrt h )

Divide both sides by 1517, so ( u + 4250) / 1517 = (1517 * sqrt h / 1517 )

​sqrt h = (u + 4250) / 1517 ,

dh = 2 * ( (u + 4250) du / 1517) / 1517 , multiply top and bottom by 1517 , so, dh = 2 (u + 4250) du / (1517^2 or 2301289)

h = (u + 4250)^2 / (1517^2 or 2301289) , h^2 = (u + 4250)^4 / 1517^4

Now using equation Integral [ (h^2 + 4h + 4) / u ] * [ (2 * sqrt h * du) / 1517 ] we can put in h in terms of u and dh in terms of u and du

Integral [ ( (u + 4250)^4 / 1517^4 ) + 4 ( (u + 4250)^2 / 2301289 ) + 4 ] / u * [ 2 (u + 4250) du / 2301289 ]

Or Integral [ ( (u + 4250)^4 / 1517^4 ) + 4 ( (u + 4250)^2 / 1517^2 ) + 4 ] / u * [ 2 (u + 4250) du / 1517^2 ]

Now we will work out the lowest common denominator for this part
[ ( (u + 4250)^4 / 1517^4 ) + 4 ( (u + 4250)^2 / 1517^2 ) + 4 ]

h^2 term has denominator 1517^4 and 4h term has 1517^2 or 2301289

So lowest common denominator of 1517^4 is with h^2 term, so this means that h term has to be multiplied by itself top and bottom to make 1517^4

For 4h term 4 ( (u + 4250)^2 / 2301289 ) we can multiply top and bottom by 1517^2 or 2301289

So, we have 4 ( 1517^2 (u + 4250)^2 / ( 2301289 * 1517^2 or 1517^4) )

and 4 term can have numerator and denominator of 1517^4

The equation now becomes Integral [ ( (u + 4250)^4 / 1517^4 ) + 4 ( 1517^2 (u + 4250)^2 / (1517^4 or 1517^2 * 1517^2) ) + ​​4 (1517^4 / 1517^4) ] / u

h part of equation has u as denominator. As there is a denominator of 1517^4 in the top part of this equation, this means that we have to multiply top and bottom for this h part of the equation by 1517^4 in an attempt to get rid of denominator on top.

Equation now becomes

Integral [ ( 1517^4 (u + 4250)^4 / 1517^4 ) + 4 ( 1517^2 * 1517^4 (u + 4250)^2 / (1517^4) ) + ​​4 (1517^4 * 1517^4 / 1517^4) ] / (1517^4 u)

1517^4 cancels on top for h^2 and 4h term

Integral [ ( (u + 4250)^4 ) + 4 ( 1517^2 (u + 4250)^2) + ​​4 (1517^4 ) ] / (1517^4 u)

Now we should expand (u + 4250)^4 and (u + 4250)^2

Now, let's expand ( u + 4250)^4 We will Use binomial theorem to expand this For the fourth power, the binomial coefficients are 1 , 4 , 6 , 4 , 1

Let's denote x = u and y = 4250 As an example, expanding for fourth power (x + y)^4 = 1 x^4 + 4x^3 *y + 6x^2 * y^2 + 4x *y^3 + 1 x^0 * 1 y^4

x^0 = 1

so, here we have (u + 4250)^4 = (1 u^4) + (4 u^3 * 4250) + (6 u^2 * 4250^2) + (4 u *4250^3) + (1 * 4250^4)

(u + 4250)^4 = u^4 + 17000 u^3 + 108375000 u^2 + 76765625000 u + 326253906250000 should be + 307062500000 u

So (u + 4250)^4 = u^4 + 17000 u^3 + 108375000 u^2 + 307062500000 u + 326253906250000

Now, let's expand ( u + 4250)^2 We will Use binomial theorem to expand this For the second power, the binomial coefficients are 1 , 2 , 1

Let's denote x = u and y = 4250 As an example, expanding for second power (x + y)^2 = (x + y)(x + y) = 1 x^2 + x *y + x * y + * 1 y^2 = x^2 + 2 x * y + y^2 x^0 = 1

so, here we have (u + 4250)^2 = (1 u^2) + (2 u * 4250) + u^2 (u + 4250)^2 = u^2 + 8500 u + 4250^2

Integral [ ( u^4 + 17000 u^3 + 108375000 u^2 + 307062500000 u + 326253906250000 ) + 4 ( 1517^2 (u^2 + 8500 u + 4250^2 ) ) + ​​4 (1517^4 ) ] / (1517^4 u)

Now from before for dh part we have dh = 2 (u + 4250) du / (1517^2 or 2301289)

Now for dh part of (2 (u + 4250) du / 1517^2 ) denominator is 1517^2 and for h^2 + 4 h + 4 part is ( 1517^4 u)

We have to multiply these denominators together (1517^4 u)(1517^2)

Now it becomes Integral [ ( u^4 + 17000 u^3 + 108375000 u^2 + 307062500000 u + 326253906250000 ) + 4 ( 1517^2 (u^2 + 8500 u + 4250^2 ) ) + ​​4 (1517^4 ) ] [ 2 (u + 4250) du ] / (1517^4 * 1517^2 u)

Now we have to get another 1517^2 on bottom denominator , so it becomes 1517^4 * 1517^4 or 5295931061521 * 1517^4 (instead of 1517^4 * 1517^2) I think Integral Calculator does it this way because that denominator of 1517^ * 1517^4 divides evenly into 1517^4 and this can help with expanding top numerator when separating terms.

The top highest numerator for the 4 terms is 1517^4 ,

so this means that 4 ( 1517^2 (u^2 + 8500 u + 4250^2 ) ) gets multiplied by 1517^2
and ( u^4 + 17000 u^3 + 108375000 u^2 + 76765625000 u + 326253906250000 ) multiplied by 1517^2 and 4 (1517^4 ) multiplied by 1517^2 along with 1517^2 in for u denominator as well, so that becomes 1517^4 * 1517^4 or 5295931061521 * 1517^4 .

Now we have Integral [ ( u^4 + 17000 u^3 + 108375000 u^2 + 307062500000 u + 326253906250000 ) 1517^2 + 4 ( 1517^2 * 1517^2 (u^2 + 8500 u + 4250^2 ) ) + ​​4 1517^2 (1517^4 ) ] [ 2 (u + 4250) du ] / (1517^4 * 1517^2 * 1517^2 u or 5295931061521 * 1517^4)

Start with ( u^4 + 17000 u^3 + 1083375000 u^2 + 3070625000000 u + 326253906250000 ) 1517^2

Lets apply this to u^4 term​
u^4 * 1517^2 = 2301289 u^4​ agrees

Lets apply this to 17000 u^3 term​ 17000 u^3 * 1517^2 = 17000 u^3 * 2301289 = 39121913000 u^3 agrees

Lets apply this to 108375000 term
Should be $108375000 \times 1517^2 / 4 = 62350548843750 u^2$ agrees with Integral Calculator great

Lets apply this to $307062500000 u$ term $= 307062500000 \times 1517^2 / 34000 = 20783516281250 u$ Agrees Integral Calculator

and lets apply this to $326253906250000$ term $32653906250000 \times 1517^2 = 750804525660156250000$ agrees with Integral Calculator

Now for $4 ( 1517^2 (u^2 + 8500 u + 4250^2 ) )$ term We multiply by 1517^2 so we have $4 (1517^4 ( u^2 + 8500 u + 4250^2 ))$

Now $4 \times 8500 u = 34000 u$

For $u^2$ we have $4 (1517^4) u^2$

and $4 \times 4250^2$

and finally $4 (1517^4 \times 1517^2

So I can find all these terms except $81455156 \times 1517^4$

So now I have worked out that 4 (4250^2 + 1517^2) 1517^4) = 4 (18062500 + 23012890) 1517^4 = 4 (20363789) 1517^4

So 4 (4250^2 + 1517^2) 1517^4) = 81455156 * 1517^4

Answer 1

For the next time, try to read what you are going to post before publishing it since your explanation is useless and impossible to follow. Also, take a look at this page for using MathJax.

The best that you can do is start by taking $u:=\sqrt{h} \implies 2u\mathrm{du}=\mathrm{d}h$ and rename your integral as $$ a\int\frac{h^2+4h+4}{b-c\sqrt{h}}\mathrm{d}h = 2a\int\frac{u^5 + 4u^3+4u}{b-cu}\mathrm{d}u, $$ where $a:=0.36$ (notice that $4a=1.44$), $b:=0.034$ and $c:=0.012136$.

You can simplify the integrand doing long division between numerator and denominator, so $$ \frac{u^5 + 4u^3+4u}{b-cu} = -\frac{u^4}{c}-\frac{bu^3}{c^2}-\frac{\left(b^2+4c^2\right)u^2}{c^3}-\frac{4b\left(b^2+2c^2\right)^2u}{c^4} -\frac{\left(b^2+2c^2\right)^2}{c^5}\left(1-\frac{b}{b-cu}\right). $$

Integrating all the terms, $$ \begin{aligned} \int&\frac{u^5 + 4u^3+4u}{b-cu}\mathrm{d}u \\ &= -\frac{1}{c}\left[\frac{u^5}{5} + \frac{bu^4}{4c} + \frac{\left(b^2+4c^2\right)u^3}{3c^2} + \frac{2b\left(b^2+2c^2\right)^2u^2}{c^3} + \frac{\left(b^2+2c^2\right)^2}{c^4}\left(u-\frac{b}{c}\log(b-cu)\right)\right] + C \\ &= -\frac{1}{c}\left[\frac{h^{5/2}}{5} + \frac{bh^2}{4c} + \frac{\left(b^2+4c^2\right)h^{3/2}}{3c^2} + \frac{2b\left(b^2+2c^2\right)^2h}{c^3} +\frac{\left(b^2+2c^2\right)^2}{c^4}\left(\sqrt{h}-\frac{b}{c}\log(b-c\sqrt{h})\right) \right] + C. \end{aligned} $$

So, your final result is (you can check it here) $$ \begin{aligned} a&\int\frac{h^2+4h+4}{b-c\sqrt{h}}\mathrm{d}h \\ &= -\frac{2a}{c}\left[\frac{h^{5/2}}{5} + \frac{bh^2}{4c} + \frac{\left(b^2+4c^2\right)h^{3/2}}{3c^2} + \frac{2b\left(b^2+2c^2\right)^2h}{c^3} +\frac{\left(b^2+2c^2\right)^2}{c^4}\left(\sqrt{h}-\frac{b}{c}\log(b-c\sqrt{h})\right) \right]. \end{aligned} $$

If you take as lower and upper bounds $h=1.96$ and $h=3$, respectively, your final result is $$ \boxed{\int_{1.96}^3 \frac{0.36h^2 + 1.44h + 1.44}{0.034 - 0.012136 \sqrt{h}} \mathrm{d}h = 514.157,} $$ as WolframAlpha returns.