Math / Calculus

QuestionThis is similar to Section 4.5 Problem 18:
Determine the indefinite integral $\int \frac{12 x^{2}}{\left(x^{3}+4\right)^{2}} d x$ by substitution. (It is recommended that you check your results by differentiation.) Use capital C for the free constant.
Answer: $\square$
Hint: Follow Example 6.

Studdy Solution

STEP 1

What is this asking? We need to find the indefinite integral of a function that looks a bit complicated, but we have a secret weapon: substitution! Watch out! Don't forget to add the constant of integration, "C", because indefinite integrals have infinitely many solutions that differ by a constant.
Also, double-check your work by differentiating your answer—it should match the original function inside the integral.

STEP 2

1. Set up the substitution
2. Rewrite the integral
3. Integrate
4. Substitute back
5. Check by differentiating

STEP 3

Let's define $u = x^3 + 4$ , the inside function of the denominator.
This is our clever substitution!

STEP 4

Now, we find the derivative of $u$ with respect to $x$ : $\frac{du}{dx} = 3x^2$ This tells us how $u$ changes when $x$ changes.

STEP 5

We can rewrite this as: $du = 3x^2 \cdot dx$ This is a crucial step for substitution!

STEP 6

Our original integral is: $\int \frac{12x^2}{(x^3 + 4)^2} dx$ We want to rewrite this entirely in terms of $u$ .

STEP 7

Notice that we have $12x^2 dx$ in the numerator, and we found $du = 3x^2 dx$ .
We can multiply both sides by $4$ to get $4 \cdot du = 4 \cdot 3x^2 dx = 12x^2 dx$ .
Perfect!

STEP 8

Now, we can substitute $u$ and $du$ into the integral: $\int \frac{12x^2}{(x^3 + 4)^2} dx = \int \frac{4 \cdot du}{u^2} = 4 \int \frac{1}{u^2} du$ Look how much simpler that looks!

STEP 9

We can rewrite the integral using a negative exponent: $4 \int u^{-2} du$ This makes it easier to apply the power rule for integration.

STEP 10

Using the power rule, we add $1$ to the exponent and divide by the new exponent: $4 \int u^{-2} du = 4 \cdot \frac{u^{-2+1}}{-2+1} + C = 4 \cdot \frac{u^{-1}}{-1} + C = -4u^{-1} + C$ Don't forget the constant of integration, $C$ !

STEP 11

Remember, we originally let $u = x^3 + 4$ .
Now, we substitute this back in: $-4u^{-1} + C = -4(x^3 + 4)^{-1} + C = \frac{-4}{x^3 + 4} + C$ This is our indefinite integral!

STEP 12

Let's differentiate our result to make sure we're right: $\frac{d}{dx} \left( \frac{-4}{x^3 + 4} + C \right) = \frac{d}{dx} \left( -4(x^3 + 4)^{-1} + C \right)$

STEP 13

Using the chain rule: $-4(-1)(x^3 + 4)^{-2} \cdot 3x^2 + 0 = \frac{12x^2}{(x^3 + 4)^2}$ This matches our original function inside the integral, so we're good to go!

STEP 14

The indefinite integral is: $\frac{-4}{x^3 + 4} + C$

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Set up the substitution
2. Rewrite the integral
3. Integrate
4. Substitute back
5. Check by differentiating

STEP 3

Let's define $u = x^3 + 4$ , the inside function of the denominator.
This is our clever substitution!

STEP 4

Now, we find the derivative of $u$ with respect to $x$ : $\frac{du}{dx} = 3x^2$ This tells us how $u$ changes when $x$ changes.

STEP 5

We can rewrite this as: $du = 3x^2 \cdot dx$ This is a crucial step for substitution!

STEP 6

Our original integral is: $\int \frac{12x^2}{(x^3 + 4)^2} dx$ We want to rewrite this entirely in terms of $u$ .

STEP 7

Notice that we have $12x^2 dx$ in the numerator, and we found $du = 3x^2 dx$ .
We can multiply both sides by $4$ to get $4 \cdot du = 4 \cdot 3x^2 dx = 12x^2 dx$ .
Perfect!

STEP 8

Now, we can substitute $u$ and $du$ into the integral: $\int \frac{12x^2}{(x^3 + 4)^2} dx = \int \frac{4 \cdot du}{u^2} = 4 \int \frac{1}{u^2} du$ Look how much simpler that looks!

STEP 9

We can rewrite the integral using a negative exponent: $4 \int u^{-2} du$ This makes it easier to apply the power rule for integration.

STEP 10

Using the power rule, we add $1$ to the exponent and divide by the new exponent: $4 \int u^{-2} du = 4 \cdot \frac{u^{-2+1}}{-2+1} + C = 4 \cdot \frac{u^{-1}}{-1} + C = -4u^{-1} + C$ Don't forget the constant of integration, $C$ !

STEP 11

Remember, we originally let $u = x^3 + 4$ .
Now, we substitute this back in: $-4u^{-1} + C = -4(x^3 + 4)^{-1} + C = \frac{-4}{x^3 + 4} + C$ This is our indefinite integral!

STEP 12

Let's differentiate our result to make sure we're right: $\frac{d}{dx} \left( \frac{-4}{x^3 + 4} + C \right) = \frac{d}{dx} \left( -4(x^3 + 4)^{-1} + C \right)$

STEP 13

Using the chain rule: $-4(-1)(x^3 + 4)^{-2} \cdot 3x^2 + 0 = \frac{12x^2}{(x^3 + 4)^2}$ This matches our original function inside the integral, so we're good to go!

STEP 14

The indefinite integral is: $\frac{-4}{x^3 + 4} + C$

Get Studdy

Studdy solves anything!

Start learning now

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

Studdy Solution

STEP 1

STEP 2

1. Set up the substitution2. Rewrite the integral3. Integrate4. Substitute back5. Check by differentiating

STEP 3

Let's **define** u=x3+4u = x^3 + 4u=x3+4, the inside function of the denominator.This is our clever substitution!

STEP 4

Now, we **find** the derivative of uuu with respect to xxx: dudx=3x2 \frac{du}{dx} = 3x^2 dxdu​=3x2 This tells us how uuu changes when xxx changes.

STEP 5

We can **rewrite** this as: du=3x2⋅dx du = 3x^2 \cdot dx du=3x2⋅dx This is a crucial step for substitution!

STEP 6

Our **original integral** is: ∫12x2(x3+4)2dx \int \frac{12x^2}{(x^3 + 4)^2} dx ∫(x3+4)212x2​dx We want to rewrite this entirely in terms of uuu.

STEP 7

Notice that we have 12x2dx12x^2 dx12x2dx in the numerator, and we found du=3x2dxdu = 3x^2 dxdu=3x2dx.We can **multiply both sides by** 444 to get 4⋅du=4⋅3x2dx=12x2dx4 \cdot du = 4 \cdot 3x^2 dx = 12x^2 dx4⋅du=4⋅3x2dx=12x2dx.Perfect!

STEP 8

Now, we can **substitute** uuu and dududu into the integral: ∫12x2(x3+4)2dx=∫4⋅duu2=4∫1u2du \int \frac{12x^2}{(x^3 + 4)^2} dx = \int \frac{4 \cdot du}{u^2} = 4 \int \frac{1}{u^2} du ∫(x3+4)212x2​dx=∫u24⋅du​=4∫u21​du Look how much simpler that looks!

STEP 9

We can **rewrite** the integral using a negative exponent: 4∫u−2du 4 \int u^{-2} du 4∫u−2du This makes it easier to apply the power rule for integration.

STEP 10

STEP 11

Remember, we originally let u=x3+4u = x^3 + 4u=x3+4.Now, we **substitute** this back in: −4u−1+C=−4(x3+4)−1+C=−4x3+4+C -4u^{-1} + C = -4(x^3 + 4)^{-1} + C = \frac{-4}{x^3 + 4} + C −4u−1+C=−4(x3+4)−1+C=x3+4−4​+C This is our indefinite integral!

STEP 12

Let's **differentiate** our result to make sure we're right: ddx(−4x3+4+C)=ddx(−4(x3+4)−1+C) \frac{d}{dx} \left( \frac{-4}{x^3 + 4} + C \right) = \frac{d}{dx} \left( -4(x^3 + 4)^{-1} + C \right) dxd​(x3+4−4​+C)=dxd​(−4(x3+4)−1+C)

STEP 13

Using the **chain rule**: −4(−1)(x3+4)−2⋅3x2+0=12x2(x3+4)2 -4(-1)(x^3 + 4)^{-2} \cdot 3x^2 + 0 = \frac{12x^2}{(x^3 + 4)^2} −4(−1)(x3+4)−2⋅3x2+0=(x3+4)212x2​ This matches our original function inside the integral, so we're good to go!

STEP 14

The indefinite integral is: −4x3+4+C \frac{-4}{x^3 + 4} + C x3+4−4​+C

Get Studdy

Studdy solves anything!

Start learning now

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

1. Set up the substitution
2. Rewrite the integral
3. Integrate
4. Substitute back
5. Check by differentiating

Let's define $u = x^3 + 4$ , the inside function of the denominator.
This is our clever substitution!

Now, we find the derivative of $u$ with respect to $x$ : $\frac{du}{dx} = 3x^2$ This tells us how $u$ changes when $x$ changes.

We can rewrite this as: $du = 3x^2 \cdot dx$ This is a crucial step for substitution!

Our original integral is: $\int \frac{12x^2}{(x^3 + 4)^2} dx$ We want to rewrite this entirely in terms of $u$ .

Notice that we have $12x^2 dx$ in the numerator, and we found $du = 3x^2 dx$ .
We can multiply both sides by $4$ to get $4 \cdot du = 4 \cdot 3x^2 dx = 12x^2 dx$ .
Perfect!

Now, we can substitute $u$ and $du$ into the integral: $\int \frac{12x^2}{(x^3 + 4)^2} dx = \int \frac{4 \cdot du}{u^2} = 4 \int \frac{1}{u^2} du$ Look how much simpler that looks!

We can rewrite the integral using a negative exponent: $4 \int u^{-2} du$ This makes it easier to apply the power rule for integration.

Remember, we originally let $u = x^3 + 4$ .
Now, we substitute this back in: $-4u^{-1} + C = -4(x^3 + 4)^{-1} + C = \frac{-4}{x^3 + 4} + C$ This is our indefinite integral!

Let's differentiate our result to make sure we're right: $\frac{d}{dx} \left( \frac{-4}{x^3 + 4} + C \right) = \frac{d}{dx} \left( -4(x^3 + 4)^{-1} + C \right)$

Using the chain rule: $-4(-1)(x^3 + 4)^{-2} \cdot 3x^2 + 0 = \frac{12x^2}{(x^3 + 4)^2}$ This matches our original function inside the integral, so we're good to go!

The indefinite integral is: $\frac{-4}{x^3 + 4} + C$