1 ) ∫ x 2 − 1 x 2 d x = I s u b s t i t u t e x = 1 sin t d x = − cos t d t sin 2 t I = ∫ 1 sin 2 t − 1 1 sin t − cos t d t sin 2 t = = ∫ 1 − sin 2 t sin 2 t ( − cos t ) d t = = − ∫ cos 2 t sin t d t = = − ∫ 1 − sin 2 t sin t d t = = ∫ ( sin t − 1 sin t ) d t = = − cos t − ln ∣ tan t 2 ∣ + c = = − cos ( arcsin 1 x ) − ln ∣ tan 1 2 arcsin 1 x ∣ + c 2 ) ∫ d x 6 x + x 2 = I s u b s t i t u t e 6 x + x 2 = t + x x = t 2 6 − 2 t d x = 12 t − 2 t 2 ( 6 − 2 t ) 2 d t x + t = − t 2 + 6 t 6 − 2 t t h e n I = ∫ 12 t − 2 t 2 ( 6 − 2 t ) 2 d t − t 2 + 6 t 6 − 2 t = = = ∫ 2 d t 6 − 2 t = − ln ∣ 6 − 2 t ∣ + c = = − ln ∣ 6 − 2 ( 6 x + x 2 − x ) ∣ + c 3 ) ∫ d x x x 4 − 4 = I s u b s t i t u t e x 2 = t t = 2 sin z d t = − cos z d z sin 2 z I = ∫ x d x x 2 x 4 − 4 = = 1 2 ∫ d x 2 x 2 x 4 − 4 = = 1 2 ∫ d t t t 2 − 4 = = 1 2 ∫ − cos z d z sin 2 z d z 2 sin z 4 sin 2 z − 4 = = − 1 4 ∫ cos z sin z d z 2 cos z sin z = = − 1 8 ∫ d z = − 1 8 z + c = − 1 8 arcsin 2 x 2 + c 4 ) ∫ x 2 d x 9 − 4 x 2 = I s u b s t i t u t e x = 3 2 sin t d x = 3 2 cos t d t I = ∫ 9 4 sin 2 t ⋅ 3 2 cos t d t 9 − 4 ⋅ 9 4 sin 2 t = = ∫ 27 8 sin 2 t cos t d t 3 cos t = = 9 8 ∫ sin 2 t d t = = 9 16 ∫ ( 1 − cos 2 t ) d t = = 9 16 ( t − 1 2 sin 2 t ) + c = = 9 16 ( arcsin 2 x 3 − 1 2 sin ( 2 arcsin 2 x 3 ) ) + c 1) \int \frac{\sqrt {x^2-1}}{x^2}dx=I\\
substitute\\
x=\frac{1}{\sin t}\\
dx=-\frac{\cos t dt}{\sin^2t}\\
I=\int\frac{\sqrt{\frac{1}{\sin^2 t}-1}}{\frac{1}{\sin t}} \frac{-\cos t dt}{\sin^2t}=\\
=\int\sqrt{\frac{1-\sin^2t}{\sin^2t}}(-\cos t)dt=\\
=-\int\frac{\cos^2t}{\sin t }dt=\\
=-\int\frac{1-\sin^2t}{\sin t}dt=\\
=\int(\sin t- \frac{1}{\sin t}) dt=\\
=-\cos t -\ln|\tan\frac{t}{2}| + c=\\
=- \cos(\arcsin \frac{1}{x}) - \ln|\tan \frac{1}{2}\arcsin \frac{1}{x}| +c\\
2) \int\frac{dx}{\sqrt{6x+x^2}}=I\\
substitute\\
\sqrt{6x+x^2}=t+x\\
x=\frac{t^2}{6-2t}\\
dx=\frac{12t-2t^2}{(6-2t)^2}dt\\
x+t=\frac{-t^2+6t}{6-2t}\\
then\\
I=\int\frac{\frac{12t-2t^2}{(6-2t)^2}dt}{\frac{-t^2+6t}{6-2t}}==\\
=\int \frac{2 dt}{6-2t}=-\ln|6-2t|+c=\\
=-\ln|6-2(\sqrt{6x+x^2}-x)|+c\\
3) \int\frac{dx}{x\sqrt{x^4-4}}=I\\
substitute\\
x^2=t\\
t=\frac{2}{\sin z}\\
dt=-\frac{\cos z dz}{\sin^2z}\\
I=\int\frac{xdx}{x^2 \sqrt{x^4-4}}=\\
=\frac{1}{2}\int\frac{dx^2}{x^2\sqrt{x^4-4}}=\\
=\frac{1}{2}\int\frac{dt}{t\sqrt{t^2-4}}=\\
=\frac{1}{2}\int\frac{-\frac{\cos z dz}{\sin^2z}dz}{\frac{2}{\sin z}\sqrt{\frac{4}{\sin^2z}-4}}=\\
=-\frac{1}{4}\int\frac{\frac{\cos z}{\sin z}dz}{\frac{2\cos z}{\sin z}}=\\
=-\frac{1}{8}\int dz=-\frac{1}{8}z+c=-\frac{1}{8}\arcsin \frac{2}{x^2}+c\\
4) \int\frac{x^2dx}{\sqrt{9-4x^2}}=I\\
substitute\\
x=\frac{3}{2}\sin t\\
dx=\frac{3}{2}\cos t dt\\
I=\int\frac{\frac{9}{4}\sin^2t \cdot\frac{3}{2}\cos t dt}{\sqrt{9-4 \cdot \frac{9}{4} \sin^2t }}=\\
=\int\frac{\frac{27}{8}\sin^2t \cos t dt}{3 \cos t}=\\
=\frac{9}{8}\int\sin^2t dt=\\
=\frac{9}{16}\int(1-\cos 2t) dt=\\
=\frac{9}{16}(t-\frac{1}{2} \sin2t) + c=\\
=\frac{9}{16}(\arcsin\frac{2x}{3}-\frac{1}{2} \sin(2\arcsin\frac{2x}{3})) + c 1 ) ∫ x 2 x 2 − 1 d x = I s u b s t i t u t e x = s i n t 1 d x = − s i n 2 t c o s t d t I = ∫ s i n t 1 s i n 2 t 1 − 1 s i n 2 t − c o s t d t = = ∫ s i n 2 t 1 − s i n 2 t ( − cos t ) d t = = − ∫ s i n t c o s 2 t d t = = − ∫ s i n t 1 − s i n 2 t d t = = ∫ ( sin t − s i n t 1 ) d t = = − cos t − ln ∣ tan 2 t ∣ + c = = − cos ( arcsin x 1 ) − ln ∣ tan 2 1 arcsin x 1 ∣ + c 2 ) ∫ 6 x + x 2 d x = I s u b s t i t u t e 6 x + x 2 = t + x x = 6 − 2 t t 2 d x = ( 6 − 2 t ) 2 12 t − 2 t 2 d t x + t = 6 − 2 t − t 2 + 6 t t h e n I = ∫ 6 − 2 t − t 2 + 6 t ( 6 − 2 t ) 2 12 t − 2 t 2 d t == = ∫ 6 − 2 t 2 d t = − ln ∣6 − 2 t ∣ + c = = − ln ∣6 − 2 ( 6 x + x 2 − x ) ∣ + c 3 ) ∫ x x 4 − 4 d x = I s u b s t i t u t e x 2 = t t = s i n z 2 d t = − s i n 2 z c o s z d z I = ∫ x 2 x 4 − 4 x d x = = 2 1 ∫ x 2 x 4 − 4 d x 2 = = 2 1 ∫ t t 2 − 4 d t = = 2 1 ∫ s i n z 2 s i n 2 z 4 − 4 − s i n 2 z c o s z d z d z = = − 4 1 ∫ s i n z 2 c o s z s i n z c o s z d z = = − 8 1 ∫ d z = − 8 1 z + c = − 8 1 arcsin x 2 2 + c 4 ) ∫ 9 − 4 x 2 x 2 d x = I s u b s t i t u t e x = 2 3 sin t d x = 2 3 cos t d t I = ∫ 9 − 4 ⋅ 4 9 s i n 2 t 4 9 s i n 2 t ⋅ 2 3 c o s t d t = = ∫ 3 c o s t 8 27 s i n 2 t c o s t d t = = 8 9 ∫ sin 2 t d t = = 16 9 ∫ ( 1 − cos 2 t ) d t = = 16 9 ( t − 2 1 sin 2 t ) + c = = 16 9 ( arcsin 3 2 x − 2 1 sin ( 2 arcsin 3 2 x )) + c
#340153 on Dec 2023