integral of sec^2(t)+t(t^2+1)^3+t^2ln(t)

解

\int sec^{2} (t) + t (t^{2} + 1)^{3} + t^{2} ln (t) d t

tan (t) + \frac{t ^{8}}{8} + \frac{t ^{6}}{2} + \frac{3 t ^{4}}{4} + \frac{t ^{2}}{2} + \frac{1}{3} t^{3} ln (t) - \frac{t ^{3}}{9} + C

解答ステップ

\int sec^{2} (t) + t (t^{2} + 1)^{3} + t^{2} ln (t) d t

総和規則を適用する:

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

= \int sec^{2} (t) d t + \int t (t^{2} + 1)^{3} d t + \int t^{2} ln (t) d t

\int sec^{2} (t) d t = tan (t)

\int t (t^{2} + 1)^{3} d t = \frac{t ^{8}}{8} + \frac{t ^{6}}{2} + \frac{3 t ^{4}}{4} + \frac{t ^{2}}{2}

\int t^{2} ln (t) d t = \frac{1}{3} t^{3} ln (t) - \frac{t ^{3}}{9}

= tan (t) + \frac{t ^{8}}{8} + \frac{t ^{6}}{2} + \frac{3 t ^{4}}{4} + \frac{t ^{2}}{2} + \frac{1}{3} t^{3} ln (t) - \frac{t ^{3}}{9}

定数を解答に追加する

= tan (t) + \frac{t ^{8}}{8} + \frac{t ^{6}}{2} + \frac{3 t ^{4}}{4} + \frac{t ^{2}}{2} + \frac{1}{3} t^{3} ln (t) - \frac{t ^{3}}{9} + C