Jobs & Education
Calculus

# What is the indefinite integral of 2x ln2x dx?

###### Wiki User

In order to work out this problem, we need to learn how to apply the integration method correctly.

The given expression is ∫ 2xln(2x) dx.

Instead of working out with 2x's, we let u = 2x. Then, du = 2 dx or du/2 = dx. This method is both valid and easy to avoid working out with too much expressions. You should get:

∫ uln(u) (du/2)

= ½ ∫ uln(u) du

Use integration by parts, which states that:

∫ f(dg) = fg - ∫ g(df)

We let:

• f = ln(u). Then, df = 1/u du
• dg = u du. Then, g = ∫ u du = ½u²

Using these substitutions, we now have:

½(½u²ln(u) - ½∫ u du)

= ¼(u²ln(u) - ∫ u du)

Finally, by integration, we obtain:

¼ * (u²ln(u) - ½u²) + c

= 1/8 * (2u²ln(u) - u²) + c

= 1/8 * (2(2x)²ln(u) - (2x)²) + c

= 1/8 * (2x)² * (2ln(u) - 1) + c

= ½ * x² * (2ln(2x - 1)) + c

🙏
0
🤨
0
😮
0
😂
0

## Related Questions

Derivative of lnx= (1/x)*(derivative of x) example: Find derivative of ln2x d(ln2x)/dx = (1/2x)*d(2x)/dx = (1/2x)*2===&gt;1/x When the problem is like ln2x^2 or ln-square root of x...., the answer won't come out in form of 1/x.

Let y = ln (2x), let u = 2x. Then:dy/dx = dy/du * du/dx= 1/(2x) * 2= 1/x.

An indefinite integral is a version of an integral that, unlike a definite integral, returns an expression instead of a number. The general form of a definite integral is: &int;ba f(x) dx. The general form of an indefinite integral is: &int; f(x) dx. An example of a definite integral is: &int;20 x2 dx. An example of an indefinite integral is: &int; x2 dx In the definite case, the answer is 23/3 - 03/3 = 8/3. In the indefinite case, the answer is x3/3 + C, where C is an arbitrary constant.

I will denote an integral as \int (LaTeX). We can let u = 2x and du = 2dx, and substitute \int ln (2x) dx = (1/2) \int ln u du. Either using integration by parts or by memorization, this is equal to (1/2) u ln u - u + C = (1/2)(2x ln (2x) - 2x) + C, where C is an arbitrary constant.

Integral of 2x dx /(2x-2) Let 2x=u 2 dx = du dx = (1/2) du Integral of 2x dx /(2x-2) = Integral of (1/2) u du / (u-2) = Integral of 1/2 [ (u-2+2) / (u-2)] dx = Integral of 1/2 [ 1+ 2/(u-2)] dx = u/2 + (1/2) 2 ln(u-2) + C = u/2 + ln(u-2) + C = (2x/2) + ln(2x-2) + C = x + ln(2x-2) + C

Integral( sin(2x)dx) = -(cos(2x)/2) + C

Integrate by parts: &int; uv dx = u &int; v dx - &int; (u' &int; v dx) dx Let u = -2x Let v = cos 3x &rarr; u' = d/dx -2x = -2 &rarr; &int; -2x cos 3x dx = -2x &int; cos 3x dx - &int; (-2 &int; cos 3x dx) dx = -2x/3 sin 3x - &int; -2/3 sin 3x dx = -2x/3 sin 3x - 2/9 cos 3x + c

convert tan^2x into sin^2x/cos^2x and secant x into 1/cos x combine terms for integral sin^2x/cos^3x dx then sub in u= cos^3x and du=-2sin^2x dx

The indefinite integral of (1/x^2)*dx is -1/x+C.

To find the integral, you can split the integral of 3sinx-5cosx into two separate integrals. For the integral of 3sinx dx integral of sinx= -cosx you can pull the three out of the integral as a coefficient Therefore, integral of 3sinx dx= -3 cosx +C For the integral of -5cosx dx integral of cosx=sinx you can pull the -5 out of the integral as a coefficient Therefore the integral of -5cosxdx= -5sinx +C Therefore the integral of 3sinx-5cosx =-3cosx-5sinx + C

This browser is pathetic for mathematical answers but here's the best that I can do:Let u = 23x therefore du/dx = 23let dv/dx = e^(2x) therefore v = 1/2*e^(2x)then, integrating by parts,I = I(u*dv/dx) dx = u*v - I(du/dx*v) dx= 23x*(1/2)*e^(2x) - I(23*(1/2)*e^(2x) dx= 23/2*xe^(2x) - 23/2*I(e^(2x)) dx= 23/2*xe^(2x) - 23/2*(1/2)*e^(2x)= 23/2*xe^(2x) - 23/4*e^(2x)or 23/4*e^(2x)*(2x - 1)

Integrate(0-&gt;t) (2-2x) dx is the integral correct (Integrate(0-&gt;t) (2-2x) dy would be different, you must state integrate with respect to what otherwise it can be anything) So integrals preserves sum and product with constants. i.e. Integrate (2-2x) dx = Integrate 2 dx - 2integrate x dx = 2x - x^2 + C By Fundamental Theorem of Calculus, take any anti-derivative, say C = 0 would be fine, and Integral(0-&gt;t)(2-2x) dx = (2x-x^2)|(0-&gt;t) = (2t-t^2) - 0 = 2t-t^2 It is a special case of the Second Fundamental Theorem of Calculus -- integral(0-&gt;t) f(x) dx is an anti-derivative of f(x).

y = 4x2 - 2 is a parabolic function with a focal point at the location (0, -2). It's derivative can be expressed as: dy/dx = 8x It's indefinite integral can be expressed as: &int;(4x2 - 2) dx = (4x3)/3 - 2x + C

,/ 2(1 - x) dx,/ 2 - 2x dx2x - x2 ...evaluated from 0 to t gives us...2t - t2 - [2(0) - (0)2]2t - t2

Depending on whether you mean 3x - 2 or 3x-2: &int;3x - 2 dx = 3x2/2 - 2x + C &int;3x-2 dx = -3/x + C

for the integral of (2x)dx/(1+x2 ) Take (1+x2 ) as your 'u' substitution. find du, du= 2x dx use u substitution to write new integral, integral of du/u the integral of du/u= ln abs(u) + C therefore, your original problem becomes an answer with ln ln abs (1+x2) + C *abs refers to absolute value of the parentheses

I wasn't entirely sure what you meant, but if the problem was to find the integral of [sec(2x)-cos(x)+x^2]dx, then in order to get the answer you must follow a couple of steps:First you should separate the problem into three parts as you are allowed to with integration. So it becomes the integral of sec(2x) - the integral of cos(x) + the integral of x^2Then solve each part separatelyThe integral of sec(2x) is -(cos(2x)/2)The integral of cos(x) is sin(x)The integral of x^2 isLastly you must combine them together:-(cos(2x)/2) - sin(x) + (x^3)/3

Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C

y=5-2xy'(u-v)=du/dx-dv/dxy'(5-2x)=d/dx(5)-d/dx(2x)-The derivative of 5 is 0 because it's a constant.-The derivative of 2x is:d/dx(cu)=c*du/dx where c is a constant.d/dx(2x)=2*d/dx(x)y'(5-2x)=(0)-[2*d/dx(x)]-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1y'(5-2x)=-[2*(1)]y'(5-2x)=-(2)y'(5-2x)= -2

This may look very daunting at first, but working through it slowly but carefully can provide the correct solution. First, I will assume that there are no bounds of integration (i.e., this is an indefinite integral). Now, the problem can be solved. So integral of (cosx)cos(sin(x))dx can be turned into "u's", using u-substituion. Let u=sinx and du = cosxdx. We are now left with integral of cos(u)du. This is sinu. But because it is an indefinite integral the constant "C" must be added. So the final answer is integral of cosxcos(sin(x))dx=integral of cosudu=sinu+C

The integral of 2 is "who gives a \$%&amp;#." You need to know what 2 is relating to and what the 2 means. If the question is "What is the integral of 2 dx" the answer would be "2x + c," with c being a constant. If you instead wish to know what the integral of 2 dy, the answer is very different. (2y +c)

Integral of [sin 2x/(1 + cos2 x)] dx (substitute 2sinx cos x for sin 2x)= Integral of [2sin x cos x/(1 + cos2 x)] dxLet u = 1 + cos2 x, so that du = -2cos x sinx dx, and dx = du/-2sin x cos x.(Substitute u for 1 + cos x, and -2sin x cos x for dx). So we have:= Integral of [(2sin x cos x/u)(du/-2sin x cos x)] (simplify)= Integral of -du/u= -ln |u| + C (substitute 1 + cos2 x for u, and take off the absolute value symbol since the expression inside it is always positive)= -ln (1 + cos2 x) + C

The derivative of a function is the function's tangent function.So, d(y)/dx = d(2tan(2x))/dx = 2*d(tan(2x))/dx = 2*d(tan(u))/du*du/dx, where u=2x, = 2*sec2(2x)*d(2x)/dx = 4*sec2(2x)*d(x)/dx = 4*sec2(2x)Now just make a plot for y = 4*sec2(2x) and you got your tangent function.

The integral of x cos(x) dx is cos(x) + x sin(x) + C

Ok, I know that sin 2x can be substituted out for 2 sin x cos x. so now I have the Integral of (sin x ) ( 2 sin x cos x ) dx which is 2 sin2x cos x dx If I use integration by parts with the u and dv, I find myself right back again..can you help. The book gives an answer, but I am not sure how it was achieved. the book gives 2/3 sin 2x cos x - 1/3 cos 2x sin x + C I may be making this more difficult than it is ? when you get the integral of 2 sin2x cos x dx use u substitution. u= sinx du= cosxdx. Then you'll get the integral of 2u^2 du.. .and then integrate...

###### Math and ArithmeticCalculusAlgebra

Copyright © 2021 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply.