Studdy AI Logo
Math  /  Calculus

Question(d) Determine the following indefinite integrals i) ∫(2+3x−4x3)dx\int\left(2+3 x-4 x^{3}\right) d x ii) ∫(cos⁡x+3sin⁡x)dx\int(\cos x+3 \sin x) d x iii) ∫4t−12t2+t+1dt\int \frac{4 t-1}{2 t^{2}+t+1} d t

Studdy Solution

STEP 1

Assumptions
1. We are given three indefinite integrals to solve.
2. We will use standard integration rules and techniques to find the antiderivatives.
3. The constants of integration will be denoted by C C .

STEP 2

For the first integral, we need to integrate each term separately.
∫(2+3x−4x3) dx\int \left(2 + 3x - 4x^3\right) \, dx

STEP 3

Apply the power rule of integration to each term. The power rule states that ∫xn dx=xn+1n+1+C\int x^n \, dx = \frac{x^{n+1}}{n+1} + C.

STEP 4

Integrate the constant term 22.
∫2 dx=2x+C\int 2 \, dx = 2x + C

STEP 5

Integrate the term 3x3x.
∫3x dx=3∫x dx=3(x22)+C=3x22+C\int 3x \, dx = 3 \int x \, dx = 3 \left( \frac{x^2}{2} \right) + C = \frac{3x^2}{2} + C

STEP 6

Integrate the term −4x3-4x^3.
∫−4x3 dx=−4∫x3 dx=−4(x44)+C=−x4+C\int -4x^3 \, dx = -4 \int x^3 \, dx = -4 \left( \frac{x^4}{4} \right) + C = -x^4 + C

STEP 7

Combine the results of the individual integrals.
∫(2+3x−4x3) dx=2x+3x22−x4+C\int \left(2 + 3x - 4x^3\right) \, dx = 2x + \frac{3x^2}{2} - x^4 + C

STEP 8

For the second integral, we need to integrate each trigonometric function separately.
∫(cos⁡x+3sin⁡x) dx\int (\cos x + 3 \sin x) \, dx

STEP 9

Integrate the term cos⁡x\cos x.
∫cos⁡x dx=sin⁡x+C\int \cos x \, dx = \sin x + C

STEP 10

Integrate the term 3sin⁡x3 \sin x.
∫3sin⁡x dx=3∫sin⁡x dx=3(−cos⁡x)+C=−3cos⁡x+C\int 3 \sin x \, dx = 3 \int \sin x \, dx = 3 \left( -\cos x \right) + C = -3 \cos x + C

STEP 11

Combine the results of the individual integrals.
∫(cos⁡x+3sin⁡x) dx=sin⁡x−3cos⁡x+C\int (\cos x + 3 \sin x) \, dx = \sin x - 3 \cos x + C

STEP 12

For the third integral, we will use the method of substitution. Let u=2t2+t+1 u = 2t^2 + t + 1 .
∫4t−12t2+t+1 dt\int \frac{4t - 1}{2t^2 + t + 1} \, dt

STEP 13

Differentiate u u with respect to t t .
du=(4t+1) dtdu = (4t + 1) \, dt

STEP 14

Rewrite the integral in terms of u u .
∫4t−1u dt\int \frac{4t - 1}{u} \, dt

STEP 15

Express dt dt in terms of du du .
dt=du4t+1dt = \frac{du}{4t + 1}

STEP 16

Substitute dt dt in the integral.
∫4t−1u⋅du4t+1\int \frac{4t - 1}{u} \cdot \frac{du}{4t + 1}

STEP 17

Simplify the integral.
∫4t−1u⋅du4t+1=∫4t−1u⋅du4t+1=∫4t−14t+1⋅duu\int \frac{4t - 1}{u} \cdot \frac{du}{4t + 1} = \int \frac{4t - 1}{u} \cdot \frac{du}{4t + 1} = \int \frac{4t - 1}{4t + 1} \cdot \frac{du}{u}

STEP 18

Recognize that the integral simplifies to a form that can be integrated directly.
∫4t−14t+1⋅duu=∫duu\int \frac{4t - 1}{4t + 1} \cdot \frac{du}{u} = \int \frac{du}{u}

STEP 19

Integrate the simplified form.
∫duu=ln⁡∣u∣+C\int \frac{du}{u} = \ln |u| + C

STEP 20

Substitute back u=2t2+t+1 u = 2t^2 + t + 1 .
ln⁡∣2t2+t+1∣+C\ln |2t^2 + t + 1| + C

STEP 21

Combine all the results for the final solutions.
i) ∫(2+3x−4x3) dx=2x+3x22−x4+C\int \left(2 + 3x - 4x^3\right) \, dx = 2x + \frac{3x^2}{2} - x^4 + C
ii) ∫(cos⁡x+3sin⁡x) dx=sin⁡x−3cos⁡x+C\int (\cos x + 3 \sin x) \, dx = \sin x - 3 \cos x + C
iii) ∫4t−12t2+t+1 dt=ln⁡∣2t2+t+1∣+C\int \frac{4t - 1}{2t^2 + t + 1} \, dt = \ln |2t^2 + t + 1| + C

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord