What is the integral of 7/((x^2+1)^{3/2)} ?

The integral of 7/((x^2+1)^{3/2)} is (7x)/((1+x^2)^{1/2)}+C

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integral of 7/((x^2+1)^{3/2)}

\int \frac{7}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\frac{7 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{7}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int cos (u) d u

= 7 \cdot \int cos (u) d u

Use the common integral:

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

= 7 sin (u)

Substitute back

u = arctan (x)

= 7 sin (arctan (x))

Simplify

7 sin (arctan (x)) : \frac{7 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{7 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{7 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

\int_{0}^{4} \frac{x}{x ^{2} - 9} d x

\int \frac{5 x + 4}{x - 1} d x

\frac{d ^{2}}{d x ^{2}} (5 e^{x} cos (x))

x \frac{d y}{d x} + x y = 1 - y, y (1) = 0

(e^{x} + y) d x + (2 + x + y e^{y}) d y = 0, y (0) = 0

What is the integral of 7/((x^2+1)^{3/2)} ?
The integral of 7/((x^2+1)^{3/2)} is (7x)/((1+x^2)^{1/2)}+C