What is the integral of 6/((t^2+1)^{3/2)} ?

The integral of 6/((t^2+1)^{3/2)} is (6t)/((1+t^2)^{1/2)}+C

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integral of 6/((t^2+1)^{3/2)}

\int \frac{6}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

\frac{6 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{6}{( 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

= 6 \cdot \int \frac{1}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

Apply Trigonometric Substitution

= 6 \cdot \int cos (u) d u

Use the common integral:

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

= 6 sin (u)

Substitute back

u = arctan (t)

= 6 sin (arctan (t))

Simplify

6 sin (arctan (t)) : \frac{6 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}}

= \frac{6 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{6 t}{( 1 + t ^{2} ) ^{\frac{1}{2}}} + C

What is the integral of 6/((t^2+1)^{3/2)} ?
The integral of 6/((t^2+1)^{3/2)} is (6t)/((1+t^2)^{1/2)}+C