Question

Question 1 (16 marks) Evaluate the following integrals, where k is a constant (2 marks each): a) ? (?x + 1?(x^2 ?x) + 1?(2 + 3x)) dx b) ? [(2x^2 + 3x + 5)?(x - 2)] dx c) ? x ln(k^2 + x^2) dx d) ? x^2 e^(-kx) dx e) ?[0 to 2] [x?(3x^2 + 4)^(3/2)] dx f) ?[0 to 4] (1 + x)^(-1) [ln(x + 1)]^3 dx g) ?[0 to ln(3)] e^(2x) (1 + e^x)^(-3/2) dx h) ?[0 to 3] |3t^2 - 12| dt

          Question 1 (16 marks) Evaluate the following integrals, where k is a constant (2 marks each):
a) ? (?x + 1?(x^2 ?x) + 1?(2 + 3x)) dx
b) ? [(2x^2 + 3x + 5)?(x - 2)] dx
c) ? x ln(k^2 + x^2) dx
d) ? x^2 e^(-kx) dx
e) ?[0 to 2] [x?(3x^2 + 4)^(3/2)] dx
f) ?[0 to 4] (1 + x)^(-1) [ln(x + 1)]^3 dx
g) ?[0 to ln(3)] e^(2x) (1 + e^x)^(-3/2) dx
h) ?[0 to 3] |3t^2 - 12| dt

Question 1 (16 marks) Evaluate the following integrals, where k is a constant (2 marks each):
a) ? (?x + 1?(x^2 ?x) + 1?(2 + 3x)) dx
b) ? [(2x^2 + 3x + 5)?(x - 2)] dx
c) ? x ln(k^2 + x^2) dx
d) ? x^2 e^(-kx) dx
e) ?[0 to 2] [x?(3x^2 + 4)^(3/2)] dx
f) ?[0 to 4] (1 + x)^(-1) [ln(x + 1)]^3 dx
g) ?[0 to ln(3)] e^(2x) (1 + e^x)^(-3/2) dx
h) ?[0 to 3] |3t^2 - 12| dt

Added by Megan W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

10/16/2023

Step 1

∫(2x+3)/(2x+3x^2) dx We can factor out 2x+3 from the denominator and write the integral as: ∫(2x+3)/(2x+3x^2) dx = ∫(2x+3)/(x(2+3x)) dx Now we can use partial fractions to split the integrand into two simpler fractions: (2x+3)/(x(2+3x)) = A/x + Show more…

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