What is the integral of t^{-1/2}(t^4+2t^2-1) ?

The integral of t^{-1/2}(t^4+2t^2-1) is 2/9 t^{9/2}+4/5 t^{5/2}-2t^{1/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of t^{-1/2}(t^4+2t^2-1)

\int t^{- \frac{1}{2}} (t^{4} + 2 t^{2} - 1) d t

\frac{2}{9} t^{\frac{9}{2}} + \frac{4}{5} t^{\frac{5}{2}} - 2 t^{\frac{1}{2}} + C

Solution steps

\int t^{- \frac{1}{2}} (t^{4} + 2 t^{2} - 1) d t

Simplify

t^{- \frac{1}{2}} (t^{4} + 2 t^{2} - 1) : \frac{t ^{4} + 2 t ^{2} - 1}{t ^{\frac{1}{2}}}

= \int \frac{t ^{4} + 2 t ^{2} - 1}{t ^{\frac{1}{2}}} d t

Expand

\frac{t ^{4} + 2 t ^{2} - 1}{t ^{\frac{1}{2}}} : t^{\frac{7}{2}} + 2 t^{\frac{3}{2}} - \frac{1}{t ^{\frac{1}{2}}}

= \int t^{\frac{7}{2}} + 2 t^{\frac{3}{2}} - \frac{1}{t ^{\frac{1}{2}}} 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^{\frac{7}{2}} d t + \int 2 t^{\frac{3}{2}} d t - \int \frac{1}{t ^{\frac{1}{2}}} d t

\int t^{\frac{7}{2}} d t = \frac{2}{9} t^{\frac{9}{2}}

\int 2 t^{\frac{3}{2}} d t = \frac{4}{5} t^{\frac{5}{2}}

\int \frac{1}{t ^{\frac{1}{2}}} d t = 2 t^{\frac{1}{2}}

= \frac{2}{9} t^{\frac{9}{2}} + \frac{4}{5} t^{\frac{5}{2}} - 2 t^{\frac{1}{2}}

Add a constant to the solution

= \frac{2}{9} t^{\frac{9}{2}} + \frac{4}{5} t^{\frac{5}{2}} - 2 t^{\frac{1}{2}} + C

\int sin (s) d s

- 4 t^{2} + 16 t + 9

\frac{\partial}{\partial x} (4 x (2 + y)^{- 1})

f (x) = \frac{x}{1 + x ^{2}}, at x = 2

\int \frac{x}{x - 10} d x

What is the integral of t^{-1/2}(t^4+2t^2-1) ?
The integral of t^{-1/2}(t^4+2t^2-1) is 2/9 t^{9/2}+4/5 t^{5/2}-2t^{1/2}+C