What is the integral of 4x^3(7x^4+12)^3 ?

The integral of 4x^3(7x^4+12)^3 is 1/4 x^4(343x^{12}+2352x^8+6048x^4+6912)+C

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integral of 4x^3(7x^4+12)^3

\int 4 x^{3} (7 x^{4} + 12)^{3} d x

\frac{1}{4} x^{4} (343 x^{12} + 2352 x^{8} + 6048 x^{4} + 6912) + C

Solution steps

\int 4 x^{3} (7 x^{4} + 12)^{3} d x

Take the constant out:

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

= 4 \cdot \int x^{3} (7 x^{4} + 12)^{3} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{3}} ( 7 u ^{\frac{4}{3}} + 12 ) ^{3}}{3} d u

= 4 \cdot \int \frac{u ^{\frac{1}{3}} ( 7 u ^{\frac{4}{3}} + 12 ) ^{3}}{3} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int u^{\frac{1}{3}} (7 u^{\frac{4}{3}} + 12)^{3} d u

Expand

u^{\frac{1}{3}} (7 u^{\frac{4}{3}} + 12)^{3} : 343 u^{\frac{13}{3}} + 1764 u^{3} + 3024 u^{\frac{5}{3}} + 1728 u^{\frac{1}{3}}

= 4 \cdot \frac{1}{3} \cdot \int 343 u^{\frac{13}{3}} + 1764 u^{3} + 3024 u^{\frac{5}{3}} + 1728 u^{\frac{1}{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{3} (\int 343 u^{\frac{13}{3}} d u + \int 1764 u^{3} d u + \int 3024 u^{\frac{5}{3}} d u + \int 1728 u^{\frac{1}{3}} d u)

\int 343 u^{\frac{13}{3}} d u = \frac{1029}{16} u^{\frac{16}{3}}

\int 1764 u^{3} d u = 441 u^{4}

\int 3024 u^{\frac{5}{3}} d u = 1134 u^{\frac{8}{3}}

\int 1728 u^{\frac{1}{3}} d u = 1296 u^{\frac{4}{3}}

= 4 \cdot \frac{1}{3} (\frac{1029}{16} u^{\frac{16}{3}} + 441 u^{4} + 1134 u^{\frac{8}{3}} + 1296 u^{\frac{4}{3}})

Substitute back

u = x^{3}

= 4 \cdot \frac{1}{3} (\frac{1029}{16} (x^{3})^{\frac{16}{3}} + 441 (x^{3})^{4} + 1134 (x^{3})^{\frac{8}{3}} + 1296 (x^{3})^{\frac{4}{3}})

Simplify

4 \cdot \frac{1}{3} (\frac{1029}{16} (x^{3})^{\frac{16}{3}} + 441 (x^{3})^{4} + 1134 (x^{3})^{\frac{8}{3}} + 1296 (x^{3})^{\frac{4}{3}}) : \frac{1}{4} x^{4} (343 x^{12} + 2352 x^{8} + 6048 x^{4} + 6912)

= \frac{1}{4} x^{4} (343 x^{12} + 2352 x^{8} + 6048 x^{4} + 6912)

Add a constant to the solution

= \frac{1}{4} x^{4} (343 x^{12} + 2352 x^{8} + 6048 x^{4} + 6912) + C

(6 x^{2} + 4 x)^{2}

\frac{\partial}{\partial y} (\frac{x - y}{x ^{2} + y ^{2}})

3 e^{t}

\int csc^{2} (θ) d θ

f (θ) = cot (θ)

What is the integral of 4x^3(7x^4+12)^3 ?
The integral of 4x^3(7x^4+12)^3 is 1/4 x^4(343x^{12}+2352x^8+6048x^4+6912)+C