Integrate (2x-3)/(2x^2+5x-3)?
∫ [(2x - 3) /(2x² + 5x - 3)] dx = factor the denominator: 2x² + 5x - 3 = 2x² + 6x - x - 3 = 2x(x + 3) - (x + 3) = (2x - 1)(x + 3) thus the integrand becomes: (2x - 3) /[(2x - 1)(x + 3)] decompose it into partial fractions: (2x - 3) /[(2x - 1)(x + 3)] = A/(2x - 1) + B/(x + 3) (2x - 3) /[(2x - 1)(x + 3)] = [A(x + 3)+ B(2x - 1)] /[(2x - 1)(x + 3)] 2x - 3 = A(x + 3) + B(2x - 1) 2x - 3 = Ax + 3A + 2Bx - B 2x - 3 = (A + 2B)x + (3A - B) yielding the system: A + 2B = 2 3A - B = - 3 A = 2 - 2B 3(2 - 2B) - B = - 3 A = 2 - 2B 6 - 6B - B = - 3 A = 2 - 2B - 7B = - 3 - 6 A = 2 - 2B 7B = 9 A = 2 - 2(9/7) = 2 - (18/7) = (14 - 18)/7 = - 4/7 B = 9/7 hence: (2x - 3) /[(2x - 1)(x + 3)] = A/(2x - 1) + B/(x + 3) = (- 4/7)/(2x - 1) + (9/7)/(x + 3) thus the integral becomes: ∫ [(2x - 3) /(2x² + 5x - 3)] dx = ∫ {[(- 4/7)/(2x - 1)] + [(9/7)/(x + 3)]} dx = break it up and pull constants out (as for the first integral we pull out - 2/7 so as to get on the top the derivative of the bottom): (- 2/7) ∫ 2 dx /(2x - 1) + (9/7) ∫ [1 /(x + 3)] dx = (- 2/7) ∫ d(2x - 1) /(2x - 1) + (9/7) ln |x + 3| + C = (- 2/7) ln |2x - 1| + (9/7) ln |x + 3| + C so the answer is: ∫ [(2x - 3) /(2x² + 5x - 3)] dx = (- 2/7) ln |2x - 1| + (9/7) ln |x + 3| + C I hope it helps..
How do i integrate (3x^4 + 5x^3 - x^2 + 2x - 1)/(x^3 + 2x^2 - 3x)dx using partial fractions?
So if you look at the integral, you notice that the degree of the numerator is larger than the degree of the denominator. What you want to do is divide the numerator by the denominator. I hope you remember how to do this. After dividing, you're left with 3x-1 + [(10x^2-x-1)/(x^3+2x^2-3x)]. When you take the integral of this new polynomial, you can divide it up into two parts. Integral [3x-1] + Integral [(10x^2-x-1)/(x^3+2x^2-3x)]. Notice that the second part of the integral ^^ you can factor both the numerator and the denominator. Integral [(10x+1)(x-1)/x(x-1)(x+3)]. The (x-1) cancel each other out. Now you're left with the Integral [(10x+1)/x(x+3)]. Set [(10x+1)/x(x+3)] = A/x + B/x+3. This is because you want to split the x and x+3. Now, set 10x+1 = A(x+3) + Bx. Group all of your x terms together. 10x+1 = Ax + 3A + Bx = (A+B)x + (3A) Notice that A+B must be equal to 10 according to the left hand side of your equation. And your 3A must be equal to one. Thus, A = 1/3 and B is 29/3. Plug A and B back into integral and now you've got: Integral [1/3x] + Integral [29/3(x+3)]. Pull the 1/3 and 29/3 out of the integral. 1/3 Integral [1/x] + 29/3 Integral [1/x+3]. = 1/3 ln(x) + 29/3 ln(x+3). I hope I didn't lose you anywhere. If you have any more questions, don't hesitate to email me. Good luck!
We want to find [math]\displaystyle \int 3^{2x} dx[/math].We can transform the integral into an expression in terms of [math]e[/math] and then use [math]u[/math]-substitution to find the final integral:[math]\displaystyle\int 3^{2x} dx[/math][math]= \displaystyle \int (e^{\ln 3})^{2x} dx[/math][math]= \displaystyle \int e^{2x \ln 3}dx[/math]Letting[math]u = 2x \ln 3 \implies[/math][math]du = 2\ln3 dx \implies dx = \dfrac{du}{2\ln 3}[/math]:[math]= \displaystyle \int e^u \dfrac{du}{2 \ln 3}[/math][math]= \displaystyle \dfrac{\int e^u du}{2 \ln 3}[/math][math]= \dfrac{e^u}{2 \ln 3}[/math][math]= \dfrac{e^{2x \ln 3}}{2 \ln 3}[/math][math]\boxed{\displaystyle\int 3^{2x} dx = \dfrac{3^{2x}}{2 \ln 3}}[/math]
You need to use a u substitution. Set u = 3x^2-7du = 6xdx -> dx = du/6xPlugging in u and for dx you get 5xdu/(6x*u^2) => 5/6 * integral of du/u^2Integrate that and then plug in back for u.
How to integrate (2x-3)^3?
There is no need to use substitution for the first integral as one answerer has done. Simply integrate as normal then divide by the derivative of the bracket. 1. Integrate this expression using the above rule: ∫ (2x - 3)³ dx = (2x - 3)⁴ / [4(2)] + C ∫ (2x - 3)³ dx = (2x - 3)⁴ / 8 + C 2. Integrate this expression by substitution: ∫ (1 - 2x)sin(1 - x + x²) dx Let u = 1 - x + x², du / dx = 2x - 1 du = (2x - 1) dx -du = (1 - 2x) dx ∫ (1 - 2x)sin(1 - x + x²) dx = - ∫ sinu du ∫ (1 - 2x)sin(1 - x + x²) dx = cosu + C Since u = 1 - x + x², ∫ (1 - 2x)sin(1 - x + x²) dx = cos(1 - x + x²) + C
Integrate (5x^2 + 3x - 2)/(x^3 + 2x^2)?
5x^2+3x-2 ___________ x^3+2x^2 5x^2 + 3x - 2.....Ax + B......C ___________ =________+____ x^2(x+2)..............x^2........x+2 5x^2 + 3x - 2 = (Ax+B)(x+2) + Cx^2 5x^2 + 3x - 2 = Ax^2 + 2Ax + Bx + 2B + Cx^2 5x^2 + 3x - 2 = x^2(A + C) + x(2A +B) +2B 2B = -2 ===>B = -1 2A + B = 3====>2A - 1 = 3====>2A = 4====>A = 2 A + C = 5====>2 + C = 5 ====> C = 3 so 5x^2+3x-2......2x-1.........3 _________ =_____+____ x^2(x + 2).........x^2........x+2 ∫ 2x dx / x^2 - ∫ dx / x^2 + 3∫ dx / (x + 2) = 2 ln x + ( 1 / x ) + 3 ln(x+2) + C
Integrate (x^3-2x^2-4)/(x^3-2x^2) from 3 to 4?
a million) f(x) = x^3 + 2x^2 – 4? What f(x) ability is take the x (the type or expression it is interior the bracket) and do it it what's shown on the superb. SO 3 * f(x+2) = 3 * ((x+2)^3 +2(x+2)^2 - 4) = 3*((x+2)(x^2+4x+4) + 2(x^2+4x+4) - 4) = 3*(x^3 + 4x^2 + 4x + 2x^2 + 8x + 8 + 2x^2 + 8x + 8 - 4) = 3*(x^2 +8x^2 + 20x + 12) = 3x^3 + 24x^2 + 60x + 36 What are each and all the attainable rational zeros for f(x) = x3 + 8x + 6 For this one, use the rational zeros theorem, which says that any rational root of an equation might nicely be expressed as r = p/q, the place p is an integer element of the final (type) term and q is an integer element of the coefficient of the optimal ability of x. So right here p = 6 and q = a million P has factors a million,2,3 and 6 (and their negatives) q has in basic terms a million and -a million So we get ±a million,±2,±3,±6 For the subsequent one, we'd like the quite zeros, so we ought to element: x^3 -3x + 4x^2 - 18 =x^3 - 3x + 3x^2 - 9 + x^2 - 9 =x(x^2-3) + 3(x^2 - 3) + x^2-9 =(x+3)(x^2-3)+(x-3)(x+3) = (x+3)(x^2-3 + x - 3) = (x+3)(x^2 +x - 6) = (x+3)(x-2)(x+3) ..answer type 3
How would you integrate 2x/(x-3)^2?
Let u = x-3........................then du = dx............................and x = u + 3 Now we have ∫ (2u + 6)du / u^2 Separate the numerator: ∫ 2u du / u^2 du + ∫ 6du/u^2 ∫ 2 du/ u = 2ln(u) <----------- first part ---------- ∫ 6 du / u^2 dx = ∫ 6*(u^(-2) du = ((6 * u^(-1)) / -1) = (-6 / u) + C = Now replace u with x - 3 (we did that at the top in reverse). (2 * ln(x + 3)) - (6 / (x-3)) + C <-------------- One answer or if you want to factor out the 2,then it's 2 * (ln(x+3) - (3 / (x-3))) + C <-------------- Another answer Corrected during proofreading -------------------------------------- To Ted S, I am legally blind although correctable to 20-30. I always come back to proofread my answers, mostly because I misread problems. Of course subscripts and superscripts are out of the question for me. It was not until after I had finished the final version that I noticed your hint. Thank you anyway. Cooperation is always happily accepted. BTW, if you allowed email this would have been between the two of us and not so public. Just a thought. I'm sure that you have your reasons. .
How do I integrate (-x)/(3x^2+2x+4)?As the solution requires a combination of different techniques, this is a useful question to ask.Firstly, I will introduce the function [math]f(x) = 3x^2 + 2x + 4[/math], which will reduce the amount of typing I will have to do.Now, note that the denominator, f(x), is a polynomial of second degree and the nominator is a polynomial of first degree. We know that differentiating a polynomial of second degree results in a polynomial of first degree, this suggests that we rewrite the integral in the form:[math]\frac {af'(x) + b}{f(x)}[/math]As [math]f'(x) = 6x + 2[/math], we can thus express the integral as:[math]\int \frac {-\frac {1}{6} f'(x) + \frac {1}{3}}{f(x)} dx[/math]We can now split this integral into two parts:[math]-\frac {1}{6} \int \frac {f'(x)}{f(x)} dx[/math][math]\frac {1}{3} \int {\frac {1}{f(x)}} dx[/math]Part 1., is easy, this is simply [math]-\frac {1}{6} ln \left| f(x) \right|[/math], where ln means natural log.Part 2. involves a little more work.[math]\frac {1}{3} \int \frac {1}{f(x)} dx = \int \frac {1}{9x^2 + 6x + 12} dx[/math]We can rewrite the denominator as:[math](3x + 1)^2 + 11 = 11 \left[ \left( \frac {3x + 1}{\sqrt {11}} \right)^2 + 1 \right][/math]Moving the factor of ‘11’ from the denominator to outside the integral, we have:[math]\frac {1}{11} \int \frac {1}{\left[ \left( \frac {3x + 1}{\sqrt {11}} \right)^2 + 1 \right]} dx[/math]This denominator is in the form [math]1 + z^2[/math], which is one of the standard integrals. It is often included in formulae sheets, but it is still useful to remember it.The standard way of solving such an integral is to use the substitution [math]tan(y) = z[/math]; differentiating both sides, we have [math]sec^2(y) dy = dz.[/math] The integral thus becomes [math]\int \frac {sec^2(y)}{1 + tan^2(y)} dy = \int dy = y + c[/math],where c is the constant of integration.For our integral, we want to use:[math]tan(y) = \frac {3x + 1}{\sqrt {11}}[/math]Differentiating both sides, we have:[math]sec^2(y) dy = \frac {3}{\sqrt {11}} dx[/math]The integral can thus be written as:[math]\frac {1}{11} \int \frac {\sqrt {11}{3}} dy = \frac {1}{3 \sqrt{11}} y + c[/math]As [math]y = tan^{-1} \left( \frac {3x + 1}{\sqrt {11}} \right)[/math]our answer is thus:[math]\frac {1}{3 \sqrt{11}} tan^{-1} \left( \frac {3x + 1}{\sqrt {11}} \right) - \frac {1}{6} ln \left| 3x^2 + 2x + 4 \right| + c[/math]