What is the integral of 7/(x^2+4) ?

The integral of 7/(x^2+4) is 7/2 arctan(x/2)+C

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integral of 7/(x^2+4)

\int \frac{7}{x ^{2} + 4} d x

\frac{7}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{7}{x ^{2} + 4} 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} + 4} d x

Apply Integral Substitution

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

Take the constant out:

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

= 7 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 7 \cdot \frac{1}{2} arctan (u)

Substitute back

u = \frac{x}{2}

= 7 \cdot \frac{1}{2} arctan (\frac{x}{2})

Simplify

7 \cdot \frac{1}{2} arctan (\frac{x}{2}) : \frac{7}{2} arctan (\frac{x}{2})

= \frac{7}{2} arctan (\frac{x}{2})

Add a constant to the solution

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

t^{2} y^{^{'}} + y = sin (t)

\frac{d}{d x} (- arcsin (\frac{a ^{2} - x ^{2}}{a}))

y = x^{2}, at x = - \frac{3}{4}

\frac{d}{d x} (ln ∣ csc (x) - cot (x) ∣)

y = x^{2} - 2 x, y = 2 - x

What is the integral of 7/(x^2+4) ?
The integral of 7/(x^2+4) is 7/2 arctan(x/2)+C