# Tag Info

Accepted

• 1,555
Accepted

### Integral of $\int \sqrt{x+\sqrt{x^2-1}}\,\mathrm{d}x$

Here is a solution that is very straightforward and requires NO substitutions - It seems to be the 'simple' solution that your maths teacher is looking for. We first start with a very helpful and cool ...
• 478

### Integral of $\int \sqrt{x+\sqrt{x^2-1}}\,\mathrm{d}x$

The substitution $x=\frac12(t+t^{-1})$ mates with $\sqrt{x^2-1}$. In your case, $$\int\sqrt{\tfrac12(t+t^{-1})+\tfrac12(t-t^{-1})}\,\tfrac12(1-t^{-2})\,dt=\tfrac12\int(\sqrt t-t^{-3/2})\,dt$$ which is ...
• 3,811
1 vote

### How do I solve $\int \frac{e^{\sqrt{x}}\sqrt{e^{\sqrt{x}}-1}}{\sqrt{x}}dx$?

Start off with $e^{\sqrt{x}}=u$ so $\frac{e^{\sqrt{x}}dx}{\sqrt{x}}=2du$ so you essentially have to $$2\int\sqrt{u-1}du$$ which is equivalent to $$\frac{4}{3}(e^{\sqrt{x}}-1)^{\frac{3}{2}}+C$$
1 vote

• 3,811
Accepted

• 6,855

Accepted

### How to integrate $\int\frac{2x+7}{4+(1+x)^2}dx$ without splitting it up?

As mentioned in the comments, one has \begin{align*} \int\frac{2x + 7}{4 + (x + 1)^{2}}\mathrm{d}x & = \int\frac{2(x + 1)}{4 + ( x + 1)^{2}}\mathrm{d}x + \int\frac{5}{4 + (x + 1)^{2}}\mathrm{d}x \...
• 17.6k

Accepted

• 270k
1 vote

\begin{aligned} I & =\int \frac{1}{\sqrt{\cos x \cdot 2 \sin ^2 \frac{x}{2}}} d x \\ & =\frac{1}{2} \int \frac{1}{\sqrt{\cos x \cdot \sin \frac{x}{2}}} d x \\ & =\int \frac{d u}{\sqrt{2 ... • 23.8k 2 votes Accepted ### Find the indefinite integral of \int \frac{1}{\sqrt{\cos(x)(1 - \cos(x))}} \, dx Applying the substitutiont = \cos x, \qquad dt = -\sin x \,dx ,$$transforms the integral to an algebraic one:$$\int \frac{dx}{\sqrt{\cos x (1 - \cos x)}} = \int \frac{dt}{(1 - t) \sqrt{t + t^2}},$... • 104k 0 votes ### Integral$\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$By substituting$x=y+\dfrac\pi4(some more details here), \begin{align*} & \int \frac{dx}{\tan x+\cot x+\csc x+\sec x} \\ &= \frac12 \int \frac{2\cos^2y-1}{\sqrt2\,\cos y+1} \, dy \\ &= ... • 21.4k 1 vote ### Evaluating \int\frac{\sqrt{1-x^2}-x}{\sqrt{1-x^2}(1+x\sqrt{1-x^2})}\, dx As an alternative to the standard tangent half-angle substitution, consider its relativeu=\tan\left(\frac\pi4-\frac\theta2\right) \iff \theta=\frac\pi2-2\arctan u$$which boils down to further ... • 21.4k 0 votes ### How to integrate \int \frac{1}{3\tan(x)-7}dx?$$\begin{align}(\log(s\sin x+c\cos x))'=\frac{s\cos x-c\sin x}{s\sin x+c\cos x}=\frac{s-c\tan x}{s\tan x+c}\\\\ (x)'=\frac{s\tan x+c}{s\tan x+c}\end{align}$$From this, you can form any linear ... • 3,811 1 vote ### How to integrate \int\frac{1}{4\tan^{2}(x)+9}dx? As you can verify,$$\left(\frac pq\arctan\left(\frac pq\tan x\right)\right)'=\frac{p^2\tan^2x+p^2}{p^2\tan^2xq^2}=1+\frac{p^2-q^2}{p^2\tan^2x+q^2}.$$Hence,$$\int\frac{dx}{p^2\tan^2x+q^2}=\frac1{p^2-... • 3,811 1 vote ### How to integrate\int\frac{1}{4\tan^{2}(x)+9}dx$? In general, let$f$be a rational function. Then the integration $$\int f(\tan x)~dx$$ can be solved by using the substitution$u=\tan{x}$. This is because$du=\sec^2x~dx=(u^2+1)dx$. Therefore, the ... • 1,952 2 votes ### Why the integral of$\sin^{-1}(x)doesnt exist? By parts, $$\int\arcsin(x)\,dx=x\arcsin(x)-\int\frac{x}{\sqrt{1-x^2}}dx=x\arcsin(x)+\sqrt{1-x^2}+C.$$ $$\int\frac{dx}{\sin x}=\int\frac{\sin x\,dx}{\sin^2 x}=-\int \frac{d\cos x}{1-\cos^2x}=-\text{... • 3,811 4 votes ### Computing the antiderivative of \frac{1}{\sqrt{A^2+2 A B e^{C z}+B^2 e^{2Cz}-\beta^2}} Let’s take out all constants first and rewrite the integral as$$ \begin{aligned} I & =\frac{A \sqrt{(A-\beta)(A+\beta)\left(C^2+K^2\right)}}{|K|} \int \frac{d z}{\sqrt{A^2+2 A B e^{C z}+B^2 e^{... • 23.8k 5 votes ### How to integrate\int\frac{1}{4\tan^{2}(x)+9}dx? \begin{aligned} \int \frac{1}{4 \tan ^2 x+9} d x = & \int \frac{\sec ^2 x}{\sec ^2 x\left(4 \tan ^2 x+9\right)} d x \\ = & \int \frac{d t}{\left(1+t^2\right)\left(4 t^2+9\right)},\quad \... • 23.8k 2 votes ### How to integrate \int\frac{1}{4\tan^{2}(x)+9}dx? DefineI = \int \frac{1}{4\tan(x)^2+9}dx$$Multiply the top and bottom by \cot(x)^2$$I = \int \frac{\cot(x)^2}{4+9\cot(x)^2}dx$$Rewrite the numerator using Pythagorean identity: \cot(x)^2=\csc(... 2 votes ### How to integrate \int\frac{1}{4\tan^{2}(x)+9}dx? Hint: Split the integrand as follows$$\frac{1}{4\tan^{2}x+9} = \frac{A(1+\tan^2x)+B(4\tan^{2}x+9)}{4\tan^{2}x+9} = B+\frac{A\sec^2x}{4\tan^{2}x+9} and find A, B by matching the numerator. • 102k 0 votes ### How to solve \int\frac{1}{\sin(4x)\cos(5x)}dx? Employ the substitution \sin x=\text{sech}\ t instead after factoring the denominator as follows, along with dt=-\frac{dx}{\sin x} and \phi=\frac{\sqrt5+1}2 \begin{align} &\int\frac{1}{\... • 102k 1 vote ### How to solve \int\frac{1}{\sin(4x)\cos(5x)}dx? Too long for a comment. Interesting use of Chebyshev polynomial of the second kind U_8(x) to convert the integrand to a rational function:\int\frac{dx}{\sin4x\cos5x}=\int\frac{dx}{\sin9x-\sin x}\... • 12.5k 1 vote ### How to integrate\int \frac{1}{3\tan(x)-7}dx? HINT Here I propose another way to approach it: \begin{align*} \int\frac{1}{3\tan(x) - 7}\mathrm{d}x & = \int\frac{\cos(x)}{3\sin(x) - 7\cos(x)}\mathrm{d}x\\\\ & = \frac{1}{\sqrt{58}}\int\frac{... • 17.6k 2 votes ### How to integrate\int \frac{1}{3\tan(x)-7}dx$? Too long for a comment (you already received elegant and simple solutions) I think that, when using the tangent half-angle substitution$x=2 \tan ^{-1}(t)$, you forgot the$dx$$I=\int \frac{dx}{3\... • 270k 0 votes ### How to integrate \int \frac{1}{3\tan(x)-7}dx? Let \cos x\equiv A(3\sin x-7\cos x)+B(7\sin x+3\cos x) for some constants A and B, then solving$$ \left\{\begin{aligned} 3 A+7 B & =0 \cdots (1)\\ -7 A+3 B & =1 \cdots (2) \end{aligned}... • 23.8k 3 votes ### How to integrate\int \frac{1}{3\tan(x)-7}dx$? Hint. Let $$C=\int\frac{\cos x}{3\sin x-7\cos x}dx$$and $$S=\int\frac{\sin x}{3\sin x-7\cos x}dx$$ Now consider$3S-7C$and$3C+7S$• 34.6k 7 votes ### How to integrate$\int \frac{1}{3\tan(x)-7}dx$? In the first approach, split the integrand as $$\frac{\cos x}{3\sin x-7\cos x} =\frac{3}{58}\frac{3\cos x+7\sin x}{3\sin x-7\cos x} -\frac{7}{58}\frac{3\sin x-7\cos x}{3\sin x-7\cos x}$$ • 102k 0 votes ### Hint on solving$\int \frac{\sqrt{1- a^{2}+x^{2}}}{x^{2}(a^2-x^2)} dxSplit the integrand as follows $$\frac{\sqrt{1- a^{2}+x^{2}}}{x^{2}(a^2-x^2)} = \frac{1}{a^2\sqrt{1- a^{2}+x^{2}}}\left(\frac{1-a^2}{x^2}+\frac{1}{a^2-x^2}\right)$$ Then \int \frac{\sqrt{1- a^{2}+x^{... • 102k 1 vote ### Integration of \int \frac{2a^2+x^2 + b x\sqrt{a^2+x^2}}{(c+b x^2)\sqrt{a^2+x^2}+ x(a^2 +x^2)} dx Result: The integral evaluates to \begin{align} I=&\int \frac{2a^2+x^2 + b x\sqrt{a^2+x^2}}{(c+b x^2)\sqrt{a^2+x^2}+ x(a^2 +x^2)} dx\\ =&\ \frac{r_++d}{r_+-r_-}\ln\bigg( x+\frac{\sqrt{1+x^2}}{... • 102k 0 votes ### How to integrate \int\frac{\sqrt{16x^2-9}}{x}dx?\int \frac{\sqrt{16x^2-9}}{x}\mathrm dx=\int \frac{16x^2-9}{x\sqrt{16x^2-9}}\mathrm dx=16\int \frac{x}{\sqrt{16x^2-9}}\mathrm dx-9\int \frac1{x^2\sqrt{16-\frac9{x^2}}}\mathrm dx$$For the ... • 637 0 votes ### Evaluate \int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx Substitute x^2=t:$$\int \frac{\mathrm dt}{(1-t)\sqrt{t^2-1}}$$Now, put 1-t=\frac1{u}. Note that u=\frac1{1-x^2}<0 as there is \sqrt{x^4-1}=\sqrt{(x^2-1)(x^2+1)} in the denominator of the ... • 637 2 votes ### How to solve this integral \int \frac 1{\sqrt { \cos x \sin^3 x }} \mathrm dx$$\int \frac 1{\sqrt{\cos x\sin^3x}} \mathrm dx=\int \frac 1{\sin^2x\sqrt{\cot x}} \mathrm dx$$Now, this can be solved easily by substituting t=\cot x. • 637 0 votes ### Indefinite integral: \int \frac{\sqrt{x^2-6x+18}}{x-3}dx$$\int \frac{\sqrt{x^2-6x+18}}{x-3} dx=\int \frac{(x-3)^2+9}{(x-3)\sqrt{x^2-6x+18}} dx=\int \frac{x-3}{\sqrt{x^2-6x+18}} dx+\int \frac{9}{(x-3)\sqrt{x^2-6x+18}} dxNow, substitute x^2-6x+... • 637 3 votes Accepted ### Integration of \int \frac{2a^2+x^2 + b x\sqrt{a^2+x^2}}{(c+b x^2)\sqrt{a^2+x^2}+ x(a^2 +x^2)} dx Looking only at your final expression (i.e. making no comment on whether you took the right approach or did all the substitutions correctly), it is possible to factorise the quartic expression in the ... • 26.2k 2 votes ### Solve indefinite integral \int\frac{x^2}{1-x^2+\sqrt{1-x^2}}dx Rationalize the denominator \begin{align*} &\int\frac{x^2}{1-x^2+\sqrt{1-x^2}}\,dx\\ =& \int\frac{x^2}{\sqrt{1-x^2} (1+\sqrt{1-x^2} )}\,dx =\int\frac{1-\sqrt{1-x^2} }{\sqrt{1-x^2}}\,dx = \sin^{... • 102k 1 vote ### Best way to solve \int\frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x-1}+1} As @Dan suggested in commentsu = \sqrt{x}+\sqrt{x-1}+1 \implies x=\frac{\left(u^2-2 u+2\right)^2}{4 (u-1)^2}\implies dx=\frac{(u-2) u \left(u^2-2 u+2\right)}{2 (u-1)^3}\int\frac{\mathrm{d}... • 270k 2 votes Accepted ### Lengthy Indefinite Integral\int \frac{\sqrt{x^2-4x+3}}{x^2+x+1} dx$Split the integrand into two and, then, apply to both the 3rd Euler substitution$ t=\frac{\sqrt{(x-3)(x-1)}}{x-1} \begin{align} &\int \frac{\sqrt{x^2-4x+3}}{x^2+x+1} dx\\ =& \int \frac1{\... • 102k 0 votes ### Evaluating the integral\int \frac{x-1}{(x+1) \sqrt{x^3+x^2+x}}dx\begin{aligned} I& =2 \int \frac{y^2-1}{\left(y^2+1\right) \sqrt{y^2+y^2+1}} d y, \quad \text { where } y=\sqrt{x} \\ & =2 \int \frac{1-\frac{1}{y^2}}{\left(y+\frac{1}{y}\right) \sqrt{y^2+\... • 23.8k 2 votes ### Lengthy Indefinite Integral \int \frac{\sqrt{x^2-4x+3}}{x^2+x+1} dxI=\int \frac{\sqrt{x^2-4x+3}}{x^2+x+1}\, dx$$Using Euler substitution$$\sqrt{x^2-4x+3}=t+x \implies x=\frac{3-t^2}{2 (t+2)}\implies dx=-\frac{t^2+4 t+3}{2 (t+2)^2}I=-\int \frac{(t+1)^2 (t+3)^... • 270k 1 vote ### Integral of\int \frac{\sqrt{x^2-1}}{x}dx$$$\int \frac{\sqrt{x^2-1}}{x} dx$$ $$=\int \frac{x^2-1}{x\sqrt{x^2-1}} dx$$ $$=\int \frac{x}{\sqrt{x^2-1}} -\frac{1}{x\sqrt{x^2-1}}dx$$ $$=\sqrt{x^2-1}-\sec^{-1}x+C$$ • 637 4 votes ### Best way to solve$\int\frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x-1}+1}$Let$t^2=x-1$, then$\$ \begin{aligned}\int \frac{d x}{\sqrt{x}+\sqrt{x-1}+1} = & \int \frac{2 t d t}{\sqrt{t^2+1}+t+1} \\ = & \int \frac{2 t\left(t+1-\sqrt{t^2+1}\right)}{(t+1)^2-\left(t^2+1\...
• 23.8k

Top 50 recent answers are included