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Use substitution to find an indefinite integral with Question What is / -4x(3x2 – 1)?dx? Do...
QUESTION 6 · 1 POINT Evaluate the indefinite integral below. Lx?idir - dx x +1 Do not include the constant +C in your answer. Provide your answer below: P FEEDBACK Content attribution
Evaluate the following indefinite integral. | asin" (a)cos(2) de Do not include "+" in your answer. For example, if you found the antiderivative was 2x + C you would enter 24.
10 Use substitution to find the indefinite integral s -dx. (2x - 5)6
(1 point) Evaluate the indefinite integral. €2x sin(4x) dx = +C.
10. (16) Find each indefinite integral using u-substitution: a. x?(1–2x")*dx b. ſxcos(x2 – 1) dx
Use substitution to find the indefinite integral. Sox ex dx [ 5x² e 7x dx=
10. (16) Find each indefinite integral using u-substitution: a. (x*(1-2x°)* dx b. fxcos (x2 - 1) dx
Evaluate the indefinite integral. (Use C for the constant of integration.) (3 - 4x) dx fo Need Help? Read It Watch It Talk to a Tutor 8. [-/1 Points] DETAILS SCALCET8 5.5.010. Evaluate the indefinite integral. (Use C for the constant of integration.) I since sin(t)/1 + cos(t) dt Need Help? Read Talk to Tutor 9. [-/1 Points) DETAILS SCALCET8 5.5.013.MI. Evaluate the indefinite integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) dx...
(a) i) For ∫(4x−4)(2x^2-4x+2)^4 dx (upper boundry =1, lower =0) Make the substitution u=2x^2−4x+2, and write the integrand as a function of u, ∫(4x−4)(2x^2−4x+2)^4 dx =∫ and hence solve the integral as a function of u, and then find the exact value of the definite integral. ii) Make the substitution u=e^(3x)/6, and write the integrand as a function of u. ∫ e^(3x)dx/36+e^(6x)=∫ Hence solve the integral as a function of u, including a constant of integration c, and then write...
(1 point) Calculate the integral below by partial fractions and by using the indicated substitution. Be sure that you can show how the results you obtain are the same. 1. 2x - dx 1 x² – 64 +C. First, rewrite this with partial fractions: I ZX64 dx = | dx + dx = (Note that you should not include the +C in your entered answer, as it has been provided at the end of the expression.) Next, use the substitution...