What is the integral of (3t^2+2)/((t^3+2t+3)^5) ?

The integral of (3t^2+2)/((t^3+2t+3)^5) is -1/(4(t^3+2t+3)^4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (3t^2+2)/((t^3+2t+3)^5)

\int \frac{3 t ^{2} + 2}{( t ^{3} + 2 t + 3 ) ^{5}} d t

- \frac{1}{4 ( t ^{3} + 2 t + 3 ) ^{4}} + C

Solution steps

\int \frac{3 t ^{2} + 2}{( t ^{3} + 2 t + 3 ) ^{5}} d t

Apply u-substitution

= \int \frac{1}{u ^{5}} d u

Apply the Power Rule

= - \frac{1}{4 u ^{4}}

Substitute back

u = t^{3} + 2 t + 3

= - \frac{1}{4 ( t ^{3} + 2 t + 3 ) ^{4}}

Add a constant to the solution

= - \frac{1}{4 ( t ^{3} + 2 t + 3 ) ^{4}} + C

\frac{\partial}{\partial y} (cos (x + y))

\frac{d}{d x} (36 - x^{2})

\frac{\partial}{\partial x} (y^{4} cos (4 x))

\frac{\partial}{\partial x} (e^{(x y)})

\int sin (t) e^{t} d t

What is the integral of (3t^2+2)/((t^3+2t+3)^5) ?
The integral of (3t^2+2)/((t^3+2t+3)^5) is -1/(4(t^3+2t+3)^4)+C