integral of ln((x+1)^2)

Lösung

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

2 x ln (x + 1) - 2 (x + 1 - ln ∣ x + 1 ∣) + C

Schritte zur Lösung

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

Wende die partielle Integration an

= x ln ((x + 1)^{2}) - \int \frac{2 x}{x + 1} d x

\int \frac{2 x}{x + 1} d x = 2 (x + 1 - ln ∣ x + 1 ∣)

= x ln ((x + 1)^{2}) - 2 (x + 1 - ln ∣ x + 1 ∣)

Wende log Regel an

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

angenommen

x \geq 0

= 2 x ln (x + 1) - 2 (x + 1 - ln ∣ x + 1 ∣)

Füge eine Konstante zur Lösung hinzu

= 2 x ln (x + 1) - 2 (x + 1 - ln ∣ x + 1 ∣) + C

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

\int \frac{1}{3 5 ^{x}} d x

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\int \frac{y}{3 - y} d y

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