integral of (t^3+5t^2+7t+1)/(t^2+3t)

Soluzione

\int \frac{t ^{3} + 5 t ^{2} + 7 t + 1}{t ^{2} + 3 t} d t

\frac{t ^{2}}{2} + 2 t + \frac{1}{3} ln ∣ t ∣ + \frac{2}{3} ln ∣ t + 3 ∣ + C

Fasi della soluzione

\int \frac{t ^{3} + 5 t ^{2} + 7 t + 1}{t ^{2} + 3 t} d t

Completa il quadrato

t^{2} + 3 t : (t + \frac{3}{2})^{2} - \frac{9}{4}

= \int \frac{t ^{3} + 5 t ^{2} + 7 t + 1}{( t + \frac{3}{2} ) ^{2} - \frac{9}{4}} d t

Applicare la sostituzione u

= \int \frac{20 u ^{2} - 32 u + 4 ( u - \frac{3}{2} ) ^{3} + 7}{4 u ^{2} - 9} d u

Applicare la sostituzione u

= \int \frac{v ^{3} + 5 v ^{2} + 7 v + 1}{v ( v + 3 )} d v

Prendi la frazione parziale di

\frac{v ^{3} + 5 v ^{2} + 7 v + 1}{v ( v + 3 )} : v + 2 + \frac{1}{3 v} + \frac{2}{3 ( v + 3 )}

= \int v + 2 + \frac{1}{3 v} + \frac{2}{3 ( v + 3 )} d v

Applica la Regola della Somma:

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

= \int v d v + \int 2 d v + \int \frac{1}{3 v} d v + \int \frac{2}{3 ( v + 3 )} d v

\int v d v = \frac{v ^{2}}{2}

\int 2 d v = 2 v

\int \frac{1}{3 v} d v = \frac{1}{3} ln ∣ v ∣

\int \frac{2}{3 ( v + 3 )} d v = \frac{2}{3} ln ∣ v + 3 ∣

= \frac{v ^{2}}{2} + 2 v + \frac{1}{3} ln ∣ v ∣ + \frac{2}{3} ln ∣ v + 3 ∣

Sostituisci indietro

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

Semplificare

\frac{( t + \frac{3}{2} - \frac{3}{2} ) ^{2}}{2} + 2 (t + \frac{3}{2} - \frac{3}{2}) + \frac{1}{3} ln t + \frac{3}{2} - \frac{3}{2} + \frac{2}{3} ln t + \frac{3}{2} - \frac{3}{2} + 3 : \frac{t ^{2}}{2} + 2 t + \frac{1}{3} ln ∣ t ∣ + \frac{2}{3} ln ∣ t + 3 ∣

= \frac{t ^{2}}{2} + 2 t + \frac{1}{3} ln ∣ t ∣ + \frac{2}{3} ln ∣ t + 3 ∣

Aggiungi una costante alla soluzione

= \frac{t ^{2}}{2} + 2 t + \frac{1}{3} ln ∣ t ∣ + \frac{2}{3} ln ∣ t + 3 ∣ + C

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

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\int \frac{x ^{3} + 1}{y} d x

\int \frac{( 9 - x ^{2} ) ^{\frac{1}{2}}}{x ^{2}} d x