What is the integral of 9/((t^2+1)^{3/2)} ?

The integral of 9/((t^2+1)^{3/2)} is (9t)/((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 9/((t^2+1)^{3/2)}

\int \frac{9}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

\frac{9 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{9}{( 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

= 9 \cdot \int \frac{1}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

Apply Trigonometric Substitution

: \int cos (u) d u

= 9 \cdot \int cos (u) d u

Use the common integral:

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

= 9 sin (u)

Substitute back

u = arctan (t)

= 9 sin (arctan (t))

Simplify

9 sin (arctan (t)) : \frac{9 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}}

= \frac{9 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{9 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}} + C

y = x^{5 x}

\int_{- 1}^{1} (3 t - 5) d t

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

\int (3 x + 1) (4 - x) d x

\frac{( x - 2 )}{x + 3}

What is the integral of 9/((t^2+1)^{3/2)} ?
The integral of 9/((t^2+1)^{3/2)} is (9t)/((1+t^2)^{1/2)}+C