What is the integral from x to 2x of (at)/(bt^2+c) ?

The integral from x to 2x of (at)/(bt^2+c) is (a(ln|c+4x^2b|-ln|c+x^2b|))/(2b)

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integral from x to 2x of (at)/(bt^2+c)

\int_{x}^{2 x} \frac{a t}{b t ^{2} + c} d t

\frac{a ( ln c + 4 x ^{2} b - ln c + x ^{2} b )}{2 b}

Solution steps

\int_{x}^{2 x} \frac{a t}{b t ^{2} + c} d t

Take the constant out:

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

= a \cdot \int_{x}^{2 x} \frac{t}{c + b t ^{2}} d t

Apply u-substitution

: \int_{c + b x^{2}}^{c + 4 x^{2} b} \frac{1}{2 b u} d u

= a \cdot \int_{c + b x^{2}}^{c + 4 x^{2} b} \frac{1}{2 b u} d u

Take the constant out:

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

= a \frac{1}{2 b} \cdot \int_{c + b x^{2}}^{c + 4 x^{2} b} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= a \frac{1}{2 b} [ln ∣ u ∣]_{c + b x^{2}}^{c + 4 x^{2} b}

Compute the boundaries:

ln c + 4 x^{2} b - ln c + b x^{2}

= a \frac{1}{2 b} (ln c + 4 x^{2} b - ln c + b x^{2})

Simplify

= \frac{a ( ln c + 4 x ^{2} b - ln c + x ^{2} b )}{2 b}

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What is the integral from x to 2x of (at)/(bt^2+c) ?
The integral from x to 2x of (at)/(bt^2+c) is (a(ln|c+4x^2b|-ln|c+x^2b|))/(2b)