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Math  /  Calculus

QuestionEvaluate the indefinite integral by using the substitution u=x2+19\mathrm{u}=\mathrm{x}^{2}+19. ∫2x(x2+19)−10dx∫2x(x2+19)−10dx=\begin{array}{l} \int 2 x\left(x^{2}+19\right)^{-10} d x \\ \int 2 x\left(x^{2}+19\right)^{-10} d x= \end{array}

Studdy Solution

STEP 1

1. We are given the integral ∫2x(x2+19)−10 dx\int 2x(x^2 + 19)^{-10} \, dx.
2. We will use the substitution u=x2+19 u = x^2 + 19 .

STEP 2

1. Perform the substitution u=x2+19 u = x^2 + 19 .
2. Differentiate u u with respect to x x to find du du .
3. Rewrite the integral in terms of u u .
4. Integrate with respect to u u .
5. Substitute back in terms of x x .

STEP 3

Perform the substitution:
Let u=x2+19 u = x^2 + 19 .

STEP 4

Differentiate u u with respect to x x :
dudx=2x \frac{du}{dx} = 2x
Thus, du=2x dx du = 2x \, dx .

STEP 5

Rewrite the integral in terms of u u :
Since du=2x dx du = 2x \, dx , we have:
∫2x(x2+19)−10 dx=∫u−10 du \int 2x(x^2 + 19)^{-10} \, dx = \int u^{-10} \, du

STEP 6

Integrate with respect to u u :
∫u−10 du=u−9−9+C \int u^{-10} \, du = \frac{u^{-9}}{-9} + C
=−19u−9+C = -\frac{1}{9} u^{-9} + C

STEP 7

Substitute back in terms of x x :
Since u=x2+19 u = x^2 + 19 , we have:
−19(x2+19)−9+C -\frac{1}{9} (x^2 + 19)^{-9} + C
The evaluated indefinite integral is:
−19(x2+19)−9+C \boxed{-\frac{1}{9} (x^2 + 19)^{-9} + C}

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