Question Statement
Evaluate the following indefinite integrals:
i. β«(3x2β2x+1)dx
ii. β«(xβ+xβ1β)dx,x>0
iii. β«x(x+1β)dx,x>0
iv. β«(2x+3)21βdx
v. β«(xβ+1)2dx
vi. β«(xββxβ1β)2dx
vii. β«xβ3x+2βdx,x>0
viii. β«yyβ(y+1)βdy,y>0
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Background and Explanation
To solve these indefinite integrals, we need to recall some basic integration rules and strategies:
- Power Rule for Integration: For any function of the form xn, the integral is given by:
β«xndx=n+1xn+1β+C,(forΒ nξ =β1)
- Sum of Integrals: The integral of a sum of functions is the sum of their integrals:
β«(f(x)+g(x))dx=β«f(x)dx+β«g(x)dx
- Special Integrals:
- For integrals involving square roots, we use the identity xβ=x1/2.
- Use of substitution where necessary, such as when expressions can be simplified by expanding or factoring terms.
Solution
i. β«(3x2β2x+1)dx
Step-by-step:
- Split the integral:
β«(3x2β2x+1)dx=β«3x2dxββ«2xdx+β«1dx
-
Integrate each term:
- β«3x2dx=3β
3x3β=x3
- β«2xdx=2β
2x2β=x2
- β«1dx=x
-
Combine results:
x3βx2+x+Cβ
ii. β«(xβ+xβ1β)dx,x>0
Step-by-step:
- Write as powers of x:
β«(xβ+xβ1β)dx=β«x1/2dx+β«xβ1/2dx
-
Apply the power rule:
- β«x1/2dx=3/2x3/2β=32βx3/2
- β«xβ1/2dx=1/2x1/2β=2x1/2
-
Combine results:
32βx3/2+2x1/2+Cβ
iii. β«x(x+1β)dx,x>0
Step-by-step:
- Expand the integrand:
β«x(x+1β)dx=β«(x(x+1)1/2)dx=β«x3/2dx+β«xdx
-
Integrate each term:
- β«x3/2dx=52βx5/2
- β«xdx=2x2β
-
Combine results:
52βx5/2+2x2β+Cβ
iv. β«(2x+3)21βdx
Step-by-step:
- Apply substitution u=2x+3, hence du=2dx:
21ββ«(u)1/2du
-
Integrate:
- β«u1/2du=32βu3/2
-
Substitute back u=2x+3:
31β(2x+3)3/2+Cβ
v. β«(xβ+1)2dx
Step-by-step:
- Expand the integrand:
(xβ+1)2=x+2x1/2+1
-
Integrate each term:
- β«xdx=2x2β
- β«2x1/2dx=34βx3/2
- β«1dx=x
-
Combine results:
2x2β+34βx3/2+x+Cβ
vi. β«(xββxβ1β)2dx
Step-by-step:
- Expand the integrand:
(xββxβ1β)2=xβ2+x1β
-
Integrate each term:
- β«xdx=2x2β
- β«β2dx=β2x
- β«x1βdx=lnβ£xβ£
-
Combine results:
2x2ββ2x+lnβ£xβ£+Cβ
vii. β«xβ3x+2βdx,x>0
Step-by-step:
- Simplify:
β«xβ3x+2βdx=β«(3x1/2+2xβ1/2)dx
-
Integrate each term:
- β«3x1/2dx=2x3/2
- β«2xβ1/2dx=4x1/2
-
Combine results:
2x3/2+4x1/2+Cβ
viii. β«yyβ(y+1)βdy,y>0
Step-by-step:
- Simplify:
β«yyβ(y+1)βdy=β«(y1/2+yβ1/2)dy
-
Integrate each term:
- β«y1/2dy=32βy3/2
- β«yβ1/2dy=2y1/2
-
Combine results:
32βy3/2+2y1/2+Cβ
- Power Rule for Integration: β«xndx=n+1xn+1β+C
- Sum of Integrals: β«(f(x)+g(x))dx=β«f(x)dx+β«g(x)dx
Summary of Steps
- Split the integrand into manageable parts.
- Apply the power rule for each term.
- Combine the results and include the constant of integration, C.