How do you integrate $y = {(5 x + 3)}^{6}$ $y = (5x + 3)^6$ ?

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Answer 1 · 2016-11-01T01:55:53

$= \frac{1}{35} \cdot {(5 x + 3)}^{7} + C$ $=1/35*(5x+3)^7+C$

So you've got...

$\int {(5 x + 3)}^{6} d x$ $int(5x+3)^6dx$

Now, make:

$u = 5 x + 3$ $u=5x+3$

Which means that:

$\frac{d u}{d x} = 5$ $(du)/(dx)=5$

And as a result:

$\frac{d x}{d u} = \frac{1}{5}$ $(dx)/(du)=1/5$

$d x = \frac{1}{5} d u$ $dx=1/5du$

Now, transform:

$\int {(5 x + 3)}^{6} d x$ $int(5x+3)^6dx$ into $\frac{1}{5} \int u^{6} d u$ $1/5intu^6du$

Integrate:

$\frac{1}{5} \int u^{6} d u$ $1/5intu^6du$

$= \frac{1}{5} \cdot \frac{u^{7}}{7} + C$ $=1/5*u^7/7+C$

$= \frac{1}{35} \cdot u^{7} + C$ $=1/35*u^7+C$

$= \frac{1}{35} \cdot {(5 x + 3)}^{7} + C$ $=1/35*(5x+3)^7+C$

$= \int {(5 x + 3)}^{6} d x$ $=int(5x+3)^6dx$

How do you integrate y=(5x+3)6y = (5x + 3)^6?