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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
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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).
How do you evaluate the integral of 2udu divided by 1 plus u squared?
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
What is the integral of 2?
The integral of 2 is "who gives a $%&#." 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)
How do you graph a tangent function y equals 2tan 2x?
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.
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)
