What is the integral of (x^3+x)3^{x^4+2x^2} ?

The integral of (x^3+x)3^{x^4+2x^2} is (3^{x^4+2x^2})/(4ln(3))+C

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integral of (x^3+x)3^{x^4+2x^2}

\int (x^{3} + x) 3^{x^{4} + 2 x^{2}} d x

\frac{3 ^{x^{4} + 2 x^{2}}}{4 ln ( 3 )} + C

Solution steps

\int (x^{3} + x) \cdot 3^{x^{4} + 2 x^{2}} d x

Apply u-substitution

: \int \frac{3 ^{u}}{4} d u

= \int \frac{3 ^{u}}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int 3^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{1}{4} \cdot \frac{3 ^{u}}{ln ( 3 )}

Substitute back

u = x^{4} + 2 x^{2}

= \frac{1}{4} \cdot \frac{3 ^{x^{4} + 2 x^{2}}}{ln ( 3 )}

Simplify

\frac{1}{4} \cdot \frac{3 ^{x^{4} + 2 x^{2}}}{ln ( 3 )} : \frac{3 ^{x^{4} + 2 x^{2}}}{4 ln ( 3 )}

= \frac{3 ^{x^{4} + 2 x^{2}}}{4 ln ( 3 )}

Add a constant to the solution

= \frac{3 ^{x^{4} + 2 x^{2}}}{4 ln ( 3 )} + C

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What is the integral of (x^3+x)3^{x^4+2x^2} ?
The integral of (x^3+x)3^{x^4+2x^2} is (3^{x^4+2x^2})/(4ln(3))+C