What is the integral of (4t)/((t^2+3)^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4t)/((t^2+3)^2)

\int \frac{4 t}{( t ^{2} + 3 ) ^{2}} d t

- \frac{2}{t ^{2} + 3} + C

Solution steps

\int \frac{4 t}{( t ^{2} + 3 ) ^{2}} d t

Take the constant out:

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

= 4 \cdot \int \frac{t}{( t ^{2} + 3 ) ^{2}} d t

Apply u-substitution

: \int \frac{1}{2 u ^{2}} d u

= 4 \cdot \int \frac{1}{2 u ^{2}} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 4 \cdot \frac{1}{2} (- \frac{1}{u})

Substitute back

u = t^{2} + 3

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

Simplify

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

= - \frac{2}{t ^{2} + 3}

Add a constant to the solution

= - \frac{2}{t ^{2} + 3} + C

\int \frac{1}{x ( 4 + x ^{6} )} d x

\int \frac{( x ^{2} - 3 )}{( x + 1 ) ^{3}} d x

\int y^{- 3} + \frac{1}{x} d x

\int 1 + x^{2} (x^{2} + 2) d x

\int 7^{2} - x^{2} d x

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