What is the integral of 8t^3(1+t^4)^{10} ?

The integral of 8t^3(1+t^4)^{10} is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8t^3(1+t^4)^{10}

\int 8 t^{3} (1 + t^{4})^{10} d t

8 (\frac{t ^{4}}{4} + \frac{5 t ^{8}}{4} + \frac{15 t ^{12}}{4} + \frac{15 t ^{16}}{2} + \frac{21 t ^{20}}{2} + \frac{21 t ^{24}}{2} + \frac{15 t ^{28}}{2} + \frac{15 t ^{32}}{4} + \frac{5 t ^{36}}{4} + \frac{t ^{40}}{4} + \frac{t ^{44}}{44}) + C

Solution steps

\int 8 t^{3} (1 + t^{4})^{10} d t

Take the constant out:

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

= 8 \cdot \int t^{3} (1 + t^{4})^{10} d t

Expand

t^{3} (1 + t^{4})^{10} : t^{3} + 10 t^{7} + 45 t^{11} + 120 t^{15} + 210 t^{19} + 252 t^{23} + 210 t^{27} + 120 t^{31} + 45 t^{35} + 10 t^{39} + t^{43}

= 8 \cdot \int t^{3} + 10 t^{7} + 45 t^{11} + 120 t^{15} + 210 t^{19} + 252 t^{23} + 210 t^{27} + 120 t^{31} + 45 t^{35} + 10 t^{39} + t^{43} 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

= 8 (\int t^{3} d t + \int 10 t^{7} d t + \int 45 t^{11} d t + \int 120 t^{15} d t + \int 210 t^{19} d t + \int 252 t^{23} d t + \int 210 t^{27} d t + \int 120 t^{31} d t + \int 45 t^{35} d t + \int 10 t^{39} d t + \int t^{43} d t)

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

\int 10 t^{7} d t = \frac{5 t ^{8}}{4}

\int 45 t^{11} d t = \frac{15 t ^{12}}{4}

\int 120 t^{15} d t = \frac{15 t ^{16}}{2}

\int 210 t^{19} d t = \frac{21 t ^{20}}{2}

\int 252 t^{23} d t = \frac{21 t ^{24}}{2}

\int 210 t^{27} d t = \frac{15 t ^{28}}{2}

\int 120 t^{31} d t = \frac{15 t ^{32}}{4}

\int 45 t^{35} d t = \frac{5 t ^{36}}{4}

\int 10 t^{39} d t = \frac{t ^{40}}{4}

\int t^{43} d t = \frac{t ^{44}}{44}

= 8 (\frac{t ^{4}}{4} + \frac{5 t ^{8}}{4} + \frac{15 t ^{12}}{4} + \frac{15 t ^{16}}{2} + \frac{21 t ^{20}}{2} + \frac{21 t ^{24}}{2} + \frac{15 t ^{28}}{2} + \frac{15 t ^{32}}{4} + \frac{5 t ^{36}}{4} + \frac{t ^{40}}{4} + \frac{t ^{44}}{44})

Add a constant to the solution

= 8 (\frac{t ^{4}}{4} + \frac{5 t ^{8}}{4} + \frac{15 t ^{12}}{4} + \frac{15 t ^{16}}{2} + \frac{21 t ^{20}}{2} + \frac{21 t ^{24}}{2} + \frac{15 t ^{28}}{2} + \frac{15 t ^{32}}{4} + \frac{5 t ^{36}}{4} + \frac{t ^{40}}{4} + \frac{t ^{44}}{44}) + C

\int \frac{- 1}{x ( ln ( x ) ) ^{2}} d x

x^{2} + y^{2} = 17, (4, - 1)

\int ln (6 x^{3}) d x

f (x) = x \cdot e^{x}