What is the integral of (12)/((t^2+1)^{3/2)} ?

The integral of (12)/((t^2+1)^{3/2)} is (12t)/((1+t^2)^{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 (12)/((t^2+1)^{3/2)}

\int \frac{12}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

\frac{12 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{12}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

Take the constant out:

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

= 12 \cdot \int \frac{1}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

Apply Trigonometric Substitution

: \int cos (u) d u

= 12 \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= 12 sin (u)

Substitute back

u = arctan (t)

= 12 sin (arctan (t))

Simplify

12 sin (arctan (t)) : \frac{12 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}}

= \frac{12 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{12 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}} + C

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

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

\int x^{3} e^{2 x} d x

n = 1 \sum \infty \frac{n ^{5}}{2 ^{n}}

F (x) = sin (2 x) - 2 x cos (2 x)

What is the integral of (12)/((t^2+1)^{3/2)} ?
The integral of (12)/((t^2+1)^{3/2)} is (12t)/((1+t^2)^{1/2)}+C