What is the integral of 2x^3(3x^4-5) ?

The integral of 2x^3(3x^4-5) is 2(1/4 x^4(3x^4-5)-(3x^8)/8)+C

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integral of 2x^3(3x^4-5)

\int 2 x^{3} (3 x^{4} - 5) d x

2 (\frac{1}{4} x^{4} (3 x^{4} - 5) - \frac{3 x ^{8}}{8}) + C

Solution steps

\int 2 x^{3} (3 x^{4} - 5) d x

Take the constant out:

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

= 2 \cdot \int x^{3} (3 x^{4} - 5) d x

Apply Integration By Parts

: \frac{1}{4} x^{4} (3 x^{4} - 5) - \int 3 x^{7} d x

= 2 (\frac{1}{4} x^{4} (3 x^{4} - 5) - \int 3 x^{7} d x)

\int 3 x^{7} d x = \frac{3 x ^{8}}{8}

= 2 (\frac{1}{4} x^{4} (3 x^{4} - 5) - \frac{3 x ^{8}}{8})

Add a constant to the solution

= 2 (\frac{1}{4} x^{4} (3 x^{4} - 5) - \frac{3 x ^{8}}{8}) + C

\int_{0}^{2 π} sin^{2} (2 x) d x

\int \frac{1 + 3 t ^{5}}{2 t} d t

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

y^{^{''}} + 36 y = - 36 sec^{2} (6 x)

f (x) = x + 6

What is the integral of 2x^3(3x^4-5) ?
The integral of 2x^3(3x^4-5) is 2(1/4 x^4(3x^4-5)-(3x^8)/8)+C