# How did they do this?

1. Oct 20, 2013

### Superposed_Cat

How did they did they do the second line? Any help appreciated, thanks in advance.

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2. Oct 20, 2013

### Emphacy

You know the rules of integration right? Imagine differentiating $x^3$. You get $3x^2$.
Now reverse the differentiation.

So integrating $3x^2$ you first increase the power and then divide by the new power. So...

1) Raise the power: $3x^3$
2) Divide by the new power: $\frac{3}{3}x^3$ = $x^3$

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

3. Oct 20, 2013

### gopher_p

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

$\int 3x^\frac{3}{4}\ dx+\int 7x^{-5}\ dx+\int \frac{1}{6}x^{-\frac{1}{2}}\ dx=3\int x^\frac{3}{4}\ dx+7\int x^{-5}\ dx+\frac{1}{6}\int x^{-\frac{1}{2}}\ dx$ 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. Oct 20, 2013

### Superposed_Cat

Profuse thanks dear sir.