What is the integral of 1/((x^2+t^2)^{1.5)} ?

The integral of 1/((x^2+t^2)^{1.5)} is x/(t^2sqrt(t^2+x^2))+C

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integral of 1/((x^2+t^2)^{1.5)}

\int \frac{1}{( x ^{2} + t ^{2} ) ^{1.5}}

\frac{x}{t ^{2} t ^{2} + x ^{2}} + C

Solution steps

\int \frac{1}{( x ^{2} + t ^{2} ) ^{1.5}} d x

(t^{2} + x^{2})^{1.5} = (t^{2} + x^{2}) t^{2} + x^{2}

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

Apply Trigonometric Substitution

: \int \frac{1}{t ^{2} sec ( u )} d u

= \int \frac{1}{t ^{2} sec ( u )} d u

Take the constant out:

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

= \frac{1}{t ^{2}} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{t ^{2}} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{t ^{2}} sin (u)

Substitute back

u = arctan (\frac{1}{t} x)

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

Simplify

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

= \frac{x}{t ^{2} t ^{2} + x ^{2}}

Add a constant to the solution

= \frac{x}{t ^{2} t ^{2} + x ^{2}} + C

\frac{d}{d x} (1 - 2 x^{2})

f (x) = 2 (x - 4)^{2}

f (x) = 10 x - 3 x^{2}

\int \frac{x}{( 16 - x ^{2} ) ^{2}} d x

y = \frac{- 4 x}{x ^{2} + 1}, (1, - 2)

What is the integral of 1/((x^2+t^2)^{1.5)} ?
The integral of 1/((x^2+t^2)^{1.5)} is x/(t^2sqrt(t^2+x^2))+C