What is the integral of 9/(x^2sqrt(1+x^2)) ?

The integral of 9/(x^2sqrt(1+x^2)) is -(9sqrt(1+x^2))/x+C

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integral of 9/(x^2sqrt(1+x^2))

\int \frac{9}{x ^{2} 1 + x ^{2}} d x

- \frac{9 1 + x ^{2}}{x} + C

Solution steps

\int \frac{9}{x ^{2} 1 + x ^{2}} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{x ^{2} 1 + x ^{2}} d x

Apply Trigonometric Substitution

: \int cot^{2} (u) sec (u) d u

= 9 \cdot \int cot^{2} (u) sec (u) d u

Simplify

cot^{2} (u) sec (u) : cot (u) csc (u)

= 9 \cdot \int cot (u) csc (u) d u

Apply u-substitution

: \int - 1 d v

= 9 \cdot \int - 1 d v

Integral of a constant:

\int a d x = a x

= 9 (- 1) v

Substitute back

= 9 (- 1) csc (arctan (x))

Simplify

9 (- 1) csc (arctan (x)) : - \frac{9 1 + x ^{2}}{x}

= - \frac{9 1 + x ^{2}}{x}

Add a constant to the solution

= - \frac{9 1 + x ^{2}}{x} + C

\int (9 x + 2) d x

\int (4 x^{2} + 3 x^{- 2}) d x

sin^{2} (2 x)

\int x^{5} x^{6} + 19 d x

\int 2 sin (2 y) d y

What is the integral of 9/(x^2sqrt(1+x^2)) ?
The integral of 9/(x^2sqrt(1+x^2)) is -(9sqrt(1+x^2))/x+C