What is the integral of x(2x+1)e^{4x^3+3x^2-4} ?

The integral of x(2x+1)e^{4x^3+3x^2-4} is 1/6 e^{4x^3+3x^2-4}+C

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integral of x(2x+1)e^{4x^3+3x^2-4}

\int x (2 x + 1) e^{4 x^{3} + 3 x^{2} - 4} d x

\frac{1}{6} e^{4 x^{3} + 3 x^{2} - 4} + C

Solution steps

\int x (2 x + 1) e^{4 x^{3} + 3 x^{2} - 4} d x

Apply u-substitution

: \int \frac{e ^{u}}{6} d u

= \int \frac{e ^{u}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{6} e^{u}

Substitute back

u = 4 x^{3} + 3 x^{2} - 4

= \frac{1}{6} e^{4 x^{3} + 3 x^{2} - 4}

Add a constant to the solution

= \frac{1}{6} e^{4 x^{3} + 3 x^{2} - 4} + C

y^{^{'}} + y^{2} = 1

\frac{( 4 x ^{5} + 5 )}{x ^{3}}

x \frac{d y}{d x} + 6 y = 1 - \frac{4}{x}

\int sec^{7} (2 x) tan (2 x) d x

\int \frac{( x + 1 ) ^{4}}{( x ^{2} + 2 x + 2 ) ^{3}} d x

What is the integral of x(2x+1)e^{4x^3+3x^2-4} ?
The integral of x(2x+1)e^{4x^3+3x^2-4} is 1/6 e^{4x^3+3x^2-4}+C