Let u=x3+2 Then du=3x2dx the integral has a 6x2 so multiplying both sides of du=3x2 by 2 gives 2 du=6x2dx. Replacing x3+2 with u and replacing 6x2dx with 2 du then gives: ∫6x2(x3+2)99dx=∫(x3+2)99(6x2dx) =∫u99(2du) =2∫u99du Integrating using the power rule then gives 2∫u99du=2⋅u100100+C=u10050+C. Substituting x3+2 for u then gives (x3+2)10050+C

Expert Answer to "∫6x2(x3+2)99dx"

College
Calculus and Analysis
Integral Calculus
Integrals

Maiclubk

Answered question

2021-08-17

\int 6 x^{2} {(x^{3} + 2)}^{99} d x

Answer & Explanation

opsadnojD

Skilled2021-08-18Added 95 answers

Let

u = x^{3} + 2

Then

d u = 3 x^{2} d x

the integral has a

6 x^{2}

so multiplying both sides of

d u = 3 x^{2}

by 2 gives 2

d u = 6 x^{2} d x

. Replacing

x^{3} + 2

with u and replacing

6 x^{2} d x

with 2 du then gives:

\int 6 x^{2} {(x^{3} + 2)}^{99} d x = \int {(x^{3} + 2)}^{99} (6 x^{2} d x)

\int u^{99} (2 d u)

2 \int u^{99} d u

Integrating using the power rule then gives

2 \int u^{99} d u = 2 \cdot \frac{u^{100}}{100} + C = \frac{u^{100}}{50} + C .

Substituting

x^{3} + 2

for u then gives

\frac{{(x^{3} + 2)}^{100}}{50} + C

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