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

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

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integral of (7x)/(x^2+4)

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

\frac{7}{2} ln x^{2} + 4 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 7 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 4

= 7 \cdot \frac{1}{2} ln x^{2} + 4

Simplify

7 \cdot \frac{1}{2} ln x^{2} + 4 : \frac{7}{2} ln x^{2} + 4

= \frac{7}{2} ln x^{2} + 4

Add a constant to the solution

= \frac{7}{2} ln x^{2} + 4 + C

(3 cos (x))^{^{'}}

\frac{\partial}{\partial x} (x^{3} y^{2} - 2 x e^{x y})

\int cot (y) csc^{2} (y) d y

y^{^{''}} - 4 y^{^{'}} + 3 y = x

y^{^{'}} - \frac{6}{x} y = \frac{y ^{4}}{x ^{3}}

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