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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
What else can I help you with?
What is the indefinite integral?
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: ∫ba f(x) dx. The general form of an indefinite integral is: ∫ f(x) dx. An example of a definite integral is: ∫20 x2 dx. An example of an indefinite integral is: ∫ 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.
What is the integral of 2-2x with limits 0 to t?
Integrate(0->t) (2-2x) dx is the integral correct (Integrate(0->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->t)(2-2x) dx = (2x-x^2)|(0->t) = (2t-t^2) - 0 = 2t-t^2 It is a special case of the Second Fundamental Theorem of Calculus -- integral(0->t) f(x) dx is an anti-derivative of f(x).
What is the integral of cosxsinx?
The integral of ( \cos x \sin x ) can be computed using a trigonometric identity. We use the identity ( \sin(2x) = 2 \sin x \cos x ), which allows us to rewrite the integral as: [ \int \cos x \sin x , dx = \frac{1}{2} \int \sin(2x) , dx. ] Integrating ( \sin(2x) ) gives: [ \frac{-1}{2} \cos(2x) + C, ] thus the final result is: [ \int \cos x \sin x , dx = \frac{-1}{4} \cos(2x) + C. ]
What is the integral for sin2 2x cos2x?
To find the integral of ( \sin^2(2x) \cos^2(2x) ), you can use the identity ( \sin^2(a) \cos^2(a) = \frac{1}{4} \sin^2(2a) ). Thus, we have: [ \sin^2(2x) \cos^2(2x) = \frac{1}{4} \sin^2(4x). ] The integral then becomes: [ \int \sin^2(2x) \cos^2(2x) , dx = \frac{1}{4} \int \sin^2(4x) , dx. ] Using the identity ( \sin^2(a) = \frac{1 - \cos(2a)}{2} ), you can further simplify and evaluate the integral.
What is the antiderivative of sin 2x?
The antiderivative of (\sin(2x)) can be found using the substitution method. The integral is given by: [ \int \sin(2x) , dx = -\frac{1}{2} \cos(2x) + C ] where (C) is the constant of integration.
What is the derivative of lnx?
Derivative of lnx= (1/x)*(derivative of x) example: Find derivative of ln2x d(ln2x)/dx = (1/2x)*d(2x)/dx = (1/2x)*2===>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.
What is the derivative of ln2x?
Let y = ln (2x), let u = 2x. Then:dy/dx = dy/du * du/dx= 1/(2x) * 2= 1/x.
What is the indefinite integral?
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: ∫ba f(x) dx. The general form of an indefinite integral is: ∫ f(x) dx. An example of a definite integral is: ∫20 x2 dx. An example of an indefinite integral is: ∫ 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.
Integral of ln2x?
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.
How do you integrate 2x over 2x-2?
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
Integration of sin2x?
Integral( sin(2x)dx) = -(cos(2x)/2) + C
How would you evaluate the indefinite integral -2xcos3xdx?
Integrate by parts: ∫ uv dx = u ∫ v dx - ∫ (u' ∫ v dx) dx Let u = -2x Let v = cos 3x → u' = d/dx -2x = -2 → ∫ -2x cos 3x dx = -2x ∫ cos 3x dx - ∫ (-2 ∫ cos 3x dx) dx = -2x/3 sin 3x - ∫ -2/3 sin 3x dx = -2x/3 sin 3x - 2/9 cos 3x + c
Integral of tan square x secant x?
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
What is the integral of 1 divided by x squared?
The indefinite integral of (1/x^2)*dx is -1/x+C.
What is the integral of 2-2x with limits 0 to t?
Integrate(0->t) (2-2x) dx is the integral correct (Integrate(0->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->t)(2-2x) dx = (2x-x^2)|(0->t) = (2t-t^2) - 0 = 2t-t^2 It is a special case of the Second Fundamental Theorem of Calculus -- integral(0->t) f(x) dx is an anti-derivative of f(x).
What is the indefinite integral of 3sinx-5cosx?
8
What is the Integral of 23x e2x dx?
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)
