What is the integral of t(t^2+1)^3 ?

The integral of t(t^2+1)^3 is (t^8)/8+(t^6)/2+(3t^4)/4+(t^2)/2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of t(t^2+1)^3

\int t (t^{2} + 1)^{3} d t

\frac{t ^{8}}{8} + \frac{t ^{6}}{2} + \frac{3 t ^{4}}{4} + \frac{t ^{2}}{2} + C

Solution steps

\int t (t^{2} + 1)^{3} d t

Expand

t (t^{2} + 1)^{3} : t^{7} + 3 t^{5} + 3 t^{3} + t

= \int t^{7} + 3 t^{5} + 3 t^{3} + t 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^{7} d t + \int 3 t^{5} d t + \int 3 t^{3} d t + \int t d t

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

\int 3 t^{5} d t = \frac{t ^{6}}{2}

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

\int t d t = \frac{t ^{2}}{2}

= \frac{t ^{8}}{8} + \frac{t ^{6}}{2} + \frac{3 t ^{4}}{4} + \frac{t ^{2}}{2}

Add a constant to the solution

= \frac{t ^{8}}{8} + \frac{t ^{6}}{2} + \frac{3 t ^{4}}{4} + \frac{t ^{2}}{2} + C

x \to 10 lim (2^{\frac{1}{x - 10}})

y^{^{'}} = \frac{y ^{2}}{1 + y ^{2}}

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4

\frac{\partial}{\partial x} (c y^{- 3})

\int 2 - \frac{x}{2} d x

What is the integral of t(t^2+1)^3 ?
The integral of t(t^2+1)^3 is (t^8)/8+(t^6)/2+(3t^4)/4+(t^2)/2+C