integral of sqrt(a^2+b^2x^2)(x)

解

\int a^{2} + b^{2} x^{2} (x) d x

\frac{1}{2} x^{2} a^{2} + b^{2} x^{2} - \frac{( b ^{2} x ^{2} - 2 a ^{2} ) b ^{2} x ^{2} + a ^{2}}{6 b ^{2}} + C

解答ステップ

\int a^{2} + b^{2} x^{2} x d x

部分積分を適用する

= \frac{1}{2} x^{2} a^{2} + b^{2} x^{2} - \int \frac{b ^{2} x ^{3}}{2 a ^{2} + b ^{2} x ^{2}} d x

\int \frac{b ^{2} x ^{3}}{2 a ^{2} + b ^{2} x ^{2}} d x = \frac{( b ^{2} x ^{2} - 2 a ^{2} ) b ^{2} x ^{2} + a ^{2}}{6 b ^{2}}

= \frac{1}{2} x^{2} a^{2} + b^{2} x^{2} - \frac{( b ^{2} x ^{2} - 2 a ^{2} ) b ^{2} x ^{2} + a ^{2}}{6 b ^{2}}

定数を解答に追加する

= \frac{1}{2} x^{2} a^{2} + b^{2} x^{2} - \frac{( b ^{2} x ^{2} - 2 a ^{2} ) b ^{2} x ^{2} + a ^{2}}{6 b ^{2}} + C