What is the integral of t^3(1+t^4)^3 ?

The integral of t^3(1+t^4)^3 is (t^4)/4+(3t^8)/8+(t^{12})/4+(t^{16})/(16)+C

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integral of t^3(1+t^4)^3

\int t^{3} (1 + t^{4})^{3} d t

\frac{t ^{4}}{4} + \frac{3 t ^{8}}{8} + \frac{t ^{12}}{4} + \frac{t ^{16}}{16} + C

Solution steps

\int t^{3} (1 + t^{4})^{3} d t

Expand

t^{3} (1 + t^{4})^{3} : t^{3} + 3 t^{7} + 3 t^{11} + t^{15}

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

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

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

\int 3 t^{11} d t = \frac{t ^{12}}{4}

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

= \frac{t ^{4}}{4} + \frac{3 t ^{8}}{8} + \frac{t ^{12}}{4} + \frac{t ^{16}}{16}

Add a constant to the solution

= \frac{t ^{4}}{4} + \frac{3 t ^{8}}{8} + \frac{t ^{12}}{4} + \frac{t ^{16}}{16} + C

\frac{d}{d x} (2 (x - 3))

\frac{d y}{d x} = \frac{( x ^{2} y - 32 )}{16 - x ^{2}} + 2

f (x) = \frac{x ^{2}}{x + 5}, at x = 3

a q^{2}

What is the integral of t^3(1+t^4)^3 ?
The integral of t^3(1+t^4)^3 is (t^4)/4+(3t^8)/8+(t^{12})/4+(t^{16})/(16)+C