- #1

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- Thread starter Superposed_Cat
- Start date

- #1

- #2

Emphacy

- 3

- 0

Now reverse the differentiation.

So integrating [itex]3x^2[/itex] you first increase the power and then divide by the new power. So...

1) Raise the power: [itex]3x^3[/itex]

2) Divide by the new power: [itex]\frac{3}{3}x^3[/itex] = [itex]x^3[/itex]

That's all that's happening on the second line.

- #3

gopher_p

- 575

- 76

##\int 3x^\frac{3}{4}\ dx+\int 7x^{-5}\ dx+\int \frac{1}{6}x^{-\frac{1}{2}}\ dx=3\int x^\frac{3}{4}\ dx+7\int x^{-5}\ dx+\frac{1}{6}\int x^{-\frac{1}{2}}\ dx## using the constant multiple rule for integrals

##3\int x^\frac{3}{4}\ dx+7\int x^{-5}\ dx+\frac{1}{6}\int x^{-\frac{1}{2}}\ dx=3\cdot\frac{1}{\frac{3}{4}+1} x^{\frac{3}{4}+1}+7\cdot\frac{1}{-5+1}x^{-5+1}+\frac{1}{6}\cdot\frac{1}{-\frac{1}{2}+1} x^{-\frac{1}{2}+1}+c## using the integral formula ##\int x^\alpha\ dx=\frac{1}{a+1}x^{\alpha+1}+c## when ##\alpha\neq-1##

- #4

Superposed_Cat

- 384

- 5

Profuse thanks dear sir.

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