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

The integral of (1+t^3)^4 is t+t^4+(6t^7)/7+(2t^{10})/5+(t^{13})/(13)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1+t^3)^4

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

t + t^{4} + \frac{6 t ^{7}}{7} + \frac{2 t ^{10}}{5} + \frac{t ^{13}}{13} + C

Solution steps

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

Expand

(1 + t^{3})^{4} : 1 + 4 t^{3} + 6 t^{6} + 4 t^{9} + t^{12}

= \int 1 + 4 t^{3} + 6 t^{6} + 4 t^{9} + t^{12} 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 1 d t + \int 4 t^{3} d t + \int 6 t^{6} d t + \int 4 t^{9} d t + \int t^{12} d t

\int 1 d t = t

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

\int 6 t^{6} d t = \frac{6 t ^{7}}{7}

\int 4 t^{9} d t = \frac{2 t ^{10}}{5}

\int t^{12} d t = \frac{t ^{13}}{13}

= t + t^{4} + \frac{6 t ^{7}}{7} + \frac{2 t ^{10}}{5} + \frac{t ^{13}}{13}

Add a constant to the solution

= t + t^{4} + \frac{6 t ^{7}}{7} + \frac{2 t ^{10}}{5} + \frac{t ^{13}}{13} + C

\int \frac{x}{5 + 2 x ^{2}} d x

\int \frac{4 x ^{3} - 4 x}{x ^{4} - 2 x ^{2} + 5} d x

\int e^{s e c (x)} sec (x) tan (x) d x

\int 8 x cos (4 x^{2} - 5) d x

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

What is the integral of (1+t^3)^4 ?
The integral of (1+t^3)^4 is t+t^4+(6t^7)/7+(2t^{10})/5+(t^{13})/(13)+C