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شائع >

integral of (t-1)/(4t^2-4t+3)

الحلّ

\int \frac{t - 1}{4 t ^{2} - 4 t + 3} d t

الحلّ

\frac{1}{8} (ln 4 t^{2} - 4 t + 3 - 2 arctan (2 (t - \frac{1}{2}))) + C

خطوات الحلّ

\int \frac{t - 1}{4 t ^{2} - 4 t + 3} d t

4 t^{2} - 4 t + 3

إكمال لمربّع:

4 (t - \frac{1}{2})^{2} + 2

= \int \frac{t - 1}{4 ( t - \frac{1}{2} ) ^{2} + 2} d t

فعّل طريقة التكامل بالتعويض

: \int \frac{2 u - 1}{2 ( 4 u ^{2} + 2 )} d u

= \int \frac{2 u - 1}{2 ( 4 u ^{2} + 2 )} d u

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

:قم بإخراج الثابت

= \frac{1}{2} \cdot \int \frac{2 u - 1}{4 u ^{2} + 2} d u

\frac{2 u - 1}{4 u ^{2} + 2}

وسٌع:

\frac{2 u}{4 u ^{2} + 2} - \frac{1}{4 u ^{2} + 2}

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

:فعّل الميزة الخطيّة

= \frac{1}{2} (\int \frac{2 u}{4 u ^{2} + 2} d u - \int \frac{1}{4 u ^{2} + 2} d u)

\int \frac{2 u}{4 u ^{2} + 2} d u = \frac{1}{4} ln 4 u^{2} + 2

\int \frac{1}{4 u ^{2} + 2} d u = \frac{1}{2 2} arctan (2 u)

= \frac{1}{2} (\frac{1}{4} ln 4 u^{2} + 2 - \frac{1}{2 2} arctan (2 u))

u = t - \frac{1}{2}

استبدل مجددًا

= \frac{1}{2} (\frac{1}{4} ln 4 (t - \frac{1}{2})^{2} + 2 - \frac{1}{2 2} arctan (2 (t - \frac{1}{2})))

\frac{1}{2} (\frac{1}{4} ln 4 (t - \frac{1}{2})^{2} + 2 - \frac{1}{2 2} arctan (2 (t - \frac{1}{2})))

بسّط:

\frac{1}{8} (ln 4 t^{2} - 4 t + 3 - 2 arctan (2 (t - \frac{1}{2})))

= \frac{1}{8} (ln 4 t^{2} - 4 t + 3 - 2 arctan (2 (t - \frac{1}{2})))

أضف ثابت التكامل للحل

= \frac{1}{8} (ln 4 t^{2} - 4 t + 3 - 2 arctan (2 (t - \frac{1}{2}))) + C

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