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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int cos (u) d u

= \int cos (u) d u

Use the common integral:

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

= sin (u)

Substitute back

u = arctan (t)

= sin (arctan (t))

Use the following identity:

sin (arctan (x)) = \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

\frac{d y}{d x} = (x + y - 2)^{2}

\frac{d ^{2}}{d x ^{2}} (3 x - 7)

2 x^{2} + 5 x + 6, (- 2, 0)

x \to \infty lim (- 1^{x})

\frac{d}{d x} ((sin (x))^{- 1})

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