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

The integral of t^3(7+t^4)^3 is (343t^4)/4+(147t^8)/8+(7t^{12})/4+(t^{16})/(16)+C

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integral of t^3(7+t^4)^3

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

\frac{343 t ^{4}}{4} + \frac{147 t ^{8}}{8} + \frac{7 t ^{12}}{4} + \frac{t ^{16}}{16} + C

Solution steps

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

Expand

t^{3} (7 + t^{4})^{3} : 343 t^{3} + 147 t^{7} + 21 t^{11} + t^{15}

= \int 343 t^{3} + 147 t^{7} + 21 t^{11} + t^{15} d t

Apply the Sum Rule:

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

= \int 343 t^{3} d t + \int 147 t^{7} d t + \int 21 t^{11} d t + \int t^{15} d t

\int 343 t^{3} d t = \frac{343 t ^{4}}{4}

\int 147 t^{7} d t = \frac{147 t ^{8}}{8}

\int 21 t^{11} d t = \frac{7 t ^{12}}{4}

\int t^{15} d t = \frac{t ^{16}}{16}

= \frac{343 t ^{4}}{4} + \frac{147 t ^{8}}{8} + \frac{7 t ^{12}}{4} + \frac{t ^{16}}{16}

Add a constant to the solution

= \frac{343 t ^{4}}{4} + \frac{147 t ^{8}}{8} + \frac{7 t ^{12}}{4} + \frac{t ^{16}}{16} + C

x \to 1 lim ((x)^{\frac{1}{x - 1}})

\int_{0}^{4} 3 x d x

\int x^{2} 36 - x^{2} d x

n = 2 \sum \infty n^{2}

\frac{d}{d x} (a x^{2} + ∣ x ∣^{3})

What is the integral of t^3(7+t^4)^3 ?
The integral of t^3(7+t^4)^3 is (343t^4)/4+(147t^8)/8+(7t^{12})/4+(t^{16})/(16)+C