How do you integrate $\int x^{2} \cdot 3^{5 x^{3} + 1}$ $int x^2*3^(5x^3+1)$ using substitution?

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Answer 1 · 2017-02-05T03:12:33

The integral equals $\frac{3^{5 x^{3} + 1}}{15 ln 3} + C$ $3^(5x^3 + 1)/(15ln3) + C$

Let $u = 5 x^{3} + 1$ $u = 5x^3 + 1$ . Then $d u = 15 x^{2} d x \to d x = \frac{d u}{15 x^{2}}$ $du = 15x^2dx -> dx = (du)/(15x^2)$ .

$\Rightarrow \int x^{2} \cdot 3^{u} \cdot \frac{d u}{15 x^{2}}$ $=>intx^2 * 3^u * (du)/(15x^2)$

$\Rightarrow \frac{1}{15} \int 3^{u} d u$ $=>1/15int 3^u du$

Use the formula $\int a^{u} d u = \frac{a^{u}}{ln a}$ $inta^udu = a^u/lna$

$\Rightarrow \frac{1}{15} (\frac{3^{u}}{ln 3}) + C$ $=>1/15(3^u/ln3) + C$

$\Rightarrow \frac{3^{5 x^{3} + 1}}{15 ln 3} + C$ $=>3^(5x^3 + 1)/(15ln3) + C$

Hopefully this helps!

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