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

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

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

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

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

Solution steps

\int \frac{2}{( 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

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

Apply Trigonometric Substitution

: \int cos (u) d u

= 2 \cdot \int cos (u) d u

Use the common integral:

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

= 2 sin (u)

Substitute back

u = arctan (t)

= 2 sin (arctan (t))

Simplify

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

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

Add a constant to the solution

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

x \to 4 lim (0.1 x^{2} + 3 x + 2)

\frac{\partial}{\partial x} (cos (x) e^{2 x y} \cdot 2 x)

\int \frac{x ^{5}}{x} d x

\int \frac{x ^{2}}{49 + x ^{6}} d x

y - x^{2}

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